More on Real Numbers

Abstract

This chapter proceeds with the study of real numbers by presenting the notion of convergence for (infinite) sequences and series of real numbers. Among other applications, we shall introduce one of the two most important numbers of Mathematics, the number e (as the reader probably suspects, the other one is the number π, which is defined as the numerical value of the area of a circle of radius 1—see [4], for instance). We shall also present a famous result of Kronecker on dense subsets of the real line, which will find several interesting applications, here and in the coming chapters.

This chapter proceeds with the study of real numbers by presenting the notion of convergence for (infinite) sequences and series of real numbers. Among other applications, we shall introduce one of the two most important numbers of Mathematics,¹ the number e. We shall also present a famous result of Kronecker on dense subsets of the real line, which will find several interesting applications, here and in the coming chapters.

7.1 Supremum and Infimum

We begin this chapter by examining the completeness of $\mathbb{R}$ from another viewpoint, for which we need to introduce some preliminary concepts.

A nonempty subset $X \subset \mathbb{R}$ is bounded from above if there exists a real number M such that

$$\displaystyle{X \subset (-\infty,M].}$$

In this case, we also say that M is an upper bound for X. Similarly, a nonempty set $X \subset \mathbb{R}$ is bounded from below if there exists a real number m such that

$$\displaystyle{X \subset [m,+\infty ).}$$

In this case, we say that m is a lower bound for X. Finally, a nonempty set $X \subset \mathbb{R}$ is bounded if X is simultaneously bounded from above and from below.

Yet in another way, a nonempty set $X \subset \mathbb{R}$ is bounded (resp. bounded from above, bounded from below) if there exists a positive real number a such that

$$\displaystyle{-a \leq x \leq a\,\,(\text{resp.}\,\,x \leq a,\,\,x \geq -a),\,\,\forall \,x \in X.}$$

On the other hand, a nonempty set $X \subset \mathbb{R}$ which is not bounded (resp. not bounded from above, not bounded from below) is said to be unbounded (resp. unbounded from above , unbounded from below ). In this case, given any positive real number a, one can always find an element x ∈ X such that

$$\displaystyle{x\notin [-a,a]\,\,(\text{resp.}\,\,x > a,\,\,x < -a).}$$

Examples 7.1

  

• (a) The set $X =\{ 1, \frac{1} {2}, \frac{1} {3}, \frac{1} {4},\ldots \}$ is bounded. Indeed, 0 is a lower bound and 1 is an upper bound for X.

• (b) Bounded intervals (cf. Definition 1.10 and subsequent discussion) are bounded sets, in the sense of the above discussion.

Example 7.2

If a nonempty set $X \subset \mathbb{R}$ is bounded from above, then the subset Y of $\mathbb{R}$ given by Y = { −x; x ∈ X} is bounded from below, and conversely. Indeed, a real number a is an upper bound for X if and only if − a is a lower bound for Y.

In spite of its apparent obviousness, we shall list our next example as an axiom of the natural number system, known as its archimedian property , after the Greek mathematician Archimedes of Syracuse.² Some important consequences of it are collected in Problems 2, 3 and 4, page 206.

Axiom 7.3

The set $\mathbb{N}$ of natural numbers is unbounded from above.

Continuing with the development of the theory, fix a bounded from above nonempty set $X \subset \mathbb{R}$. If $M \in \mathbb{R}$ is an upper bound for X, then X ⊂ (−∞, M]. Nevertheless, it may happen that there exists M ^′ < M which still is an upper bound for X, i.e., is such that X ⊂ (−∞, M ^′]. Indeed, the condition X ⊂ (−∞, M] doesn’t guarantee that, for M ^′ < M, we have X ∩ (M ^′, M] ≠ ∅; if it happens that X ∩ (M ^′, M] = ∅, then we will have X ⊂ (−∞, M ^′], and M ^′ will be another upper bound for X, which is less than M.

On the other hand, given x ∈ X, it is obvious that no M ^′ < x is an upper bound for X, since x ∈ X ∖(−∞, M ^′], i.e., X ⊄ (−∞, M ^′]. To put in another way, the set of upper bounds of a nonempty, bounded from above set $X \subset \mathbb{R}$ is bounded from below.

As we shall see later (cf. Problem 9, page 242), the completeness of $\mathbb{R}$, as postulated in Sect. 1.3, is a consequence of the following statement, which sharpens the discussion of the two previous paragraphs and will be taken as an axiom.

Axiom 7.4

If a nonempty set $X \subset \mathbb{R}$ is bounded from above, then there exists a real number M satisfying the two following properties:

• (a) M is an upper bound for X.

• (b) If M ^′ < M, then M ^′ is not an upper bound for X.

In the hypotheses and notations of the previous axiom, a moment’s thought shows that there exists at most one real number M satisfying the properties there stated. Indeed, if two distinct real numbers M ₁ and M ₂ did the job, suppose, without loss of generality, that M ₁ < M ₂. Then, on the one hand, item (a) would guarantee that M ₁ and M ₂ are upper bounds for X; on the other, since M ₁ < M ₂, item (b) would guarantee that M ₁ is not an upper bound for X.

Yet in the hypotheses and notations of the axiom, the discussion of the last paragraph allows us to say that M is the least upper bound or supremum of X. We denote M = sup(X) or M = lub(X).

Examples 7.5

  

• (a) The set $X =\{ 1, \frac{1} {2}, \frac{1} {3}, \frac{1} {4},\ldots \}$ has 1 as an upper bound. On the other hand, since 1 ∈ X, no real number less that 1 can be an upper bound of X, so that sup(X) = 1.

• (b) If X = (1, 2) (an open interval), then 2, but no real number less than 2, is an upper bound for X. Indeed, if 1 < a < 2, then the number $\frac{1+a} {2}$ is also greater than 1 and less than 2, so that $\frac{1+a} {2} \in X$. Now, since $a < \frac{1+a} {2} \in X$, the number a cannot be an upper bound for X. Therefore, sup(X) = 2, and observe that 2 ∉ X.

If $\emptyset \neq Y \subset \mathbb{R}$ is bounded below, then one can prove, as an easy consequence of Theorem 7.4 (cf. Problem 6), that Y admits a greatest lower bound m, which is also said to be an infimum of Y. As in the case of nonempty, bounded above sets, one can easily prove that a nonempty, bounded below set Y has a unique greatest lower bound m; hence, we denote m = inf(Y ) or m = glb(Y ).

For future use, the coming results collect some useful properties of the notions of sup and inf.

Proposition 7.6

Let $X \subset \mathbb{R}$ be a nonempty, bounded above set, and M = sup (X). If $n \in \mathbb{N}$ , then there exists x ∈ X such that

$$\displaystyle{M - \frac{1} {n} < x \leq M.}$$

Proof

Since M is the least upper bound of X and $M - \frac{1} {n} < M$, the real number $M - \frac{1} {n}$ is no longer an upper bound of X. Hence, there exists x ∈ X such that $x > M - \frac{1} {n}$. However, since X ⊂ (−∞, M], we must also have x ≤ M. □ 

Problem 10 states an analogous result for the greatest lower bound of nonempty, bounded below sets.

Proposition 7.7

Let $X,Y \subset \mathbb{R}$ be nonempty sets. If x ≤ y for all x ∈ X and y ∈ Y, then X is bounded above, Y is bounded below and

$$\displaystyle{\sup (X) \leq \inf (Y ).}$$

Proof

Fix y ∈ Y arbitrarily. Since x ≤ y for all x ∈ X, the number y is an upper bound for X. Hence, X is bounded above and, letting M = sup(X) (the least upper bound of X), we have M ≤ y.

Since y ∈ Y was arbitrarily chosen, the reasoning of the previous paragraph shows that, for all y ∈ Y, we have M ≤ y. Therefore, M is a lower bound for Y, so that Y is bounded below and, letting m = infY (the greatest lower bound of Y ), we have M ≤ m. □ 

The next result gives a sufficient condition for equality to happen in the previous proposition.

Proposition 7.8

Let $X,Y \subset \mathbb{R}$ be nonempty sets, such that X is bounded above, Y is bounded below and sup X ≤ inf Y. If, for every $n \in \mathbb{N}$ , there exist x _n ∈ X and y _n ∈ Y satisfying $y_{n} - x_{n} < \frac{1} {n}$ , then sup (X) = inf (Y ).

Proof

Let M = sup(X), m = inf(Y ) and suppose that M < m. Since x ≤ M < m ≤ y for all x ∈ X and y ∈ Y, we would have y − x ≥ m − M, for all x ∈ X and y ∈ Y.

On the other hand, by choosing a natural number $n > \frac{1} {m-M}$ (which is possible, thanks to Axiom 7.3), our hypotheses would guarantee the existence of real numbers x _n  ∈ X and y _n  ∈ Y such that

$$\displaystyle{y_{n} - x_{n} < \frac{1} {n} < m - M,}$$

which is a contradiction!

Finally, since the assumption that sup(X) < inf(Y ) leads us to a contradiction, the only possibility is that sup(X) = inf(Y ). □ 

We finish this section by presenting a nontrivial (and important) application of the concept of supremum of a bounded above nonempty set. To set the stage, we recall that, in the previous chapters, we several times relied upon the existence of square roots of positive real numbers. The next result establishes existence in a more general setting.

Theorem 7.9

Given a positive real number x and a natural number n > 1, there exists a unique positive real number y such that y ⁿ = x.

Proof

Let’s consider just the case x > 1, referring the reader to Problem 1 for the case 0 < x < 1 and observing that the case x = 1 is trivial. If

$$\displaystyle{X =\{ a \in \mathbb{R};\,a \geq 0\,\,\text{and}\,\,a^{n} < x\},}$$

then X is nonempty, since 0 ∈ X. Also, X is bounded above, for, if α ≥ x + 1, then Problem 6, page 176, together with item (b) of Corollary 1.3, guarantees that α ⁿ ≥ (x + 1)ⁿ > x + 1 > x; therefore, α ∉ X and, thus, X ⊂ (−∞, x + 1).

Being nonempty and bounded above, X has a least upper bound, say y. Since 1ⁿ = 1 < x, we have 1 ∈ X and, thus, y ≥ 1. We shall show that y ⁿ = x in the following way:

• (i) If y ⁿ < x, we shall obtain a positive real number z such that y ⁿ < z ⁿ < x. Then, the first inequality will give y < z, whereas the second (with the aid of the result of Problem 6, page 176) z ∈ X. Therefore, we will get y < z ∈ X, thus contradicting the fact that y is an upper bound for X.

• (ii) If y ⁿ > x, we shall obtain a positive real number z for which x < z ⁿ < y ⁿ. Then, taking an arbitrary a ∈ X, it will follow from a ⁿ < x < z ⁿ < y ⁿ (again by the result of Problem 6, page 176) that a < z < y, so that z will be an upper bound for X which is less than y. This will contradict the fact that y is the least upper bound for X.

Once we have proved that y ⁿ < x and y ⁿ > x both lead to contradictions, the only possibility left will be y ⁿ = x.

For the proof of (i), suppose y ⁿ < x. If $z = y + \frac{1} {k}$, with $k \in \mathbb{N}$, then, once more by the result of Problem 6, page 176, we have y ⁿ < z ⁿ. We assert that z ⁿ < x for all sufficiently large k. Indeed, the binomial formula (cf. Theorem 4.20) gives

$$\displaystyle{z^{n} = \left (y + \frac{1} {k}\right )^{n} = y^{n} +\sum _{ j=1}^{n}{n\choose j} \frac{1} {k^{j}}y^{n-j};}$$

however, since ${n\choose j} < 2^{n}$, $\frac{1} {k^{j}} < \frac{1} {k}$ and y ^n−j < y ⁿ for 1 ≤ j ≤ n, it follows that

$$\displaystyle{z^{n} < y^{n} +\sum _{ j=1}^{n}\frac{2^{n}} {k} y^{n} = y^{n} + \frac{n(2y)^{n}} {k};}$$

therefore, we have z ⁿ < x provided $y^{n} + \frac{n(2y)^{n}} {k} < x$, i.e., $k > \frac{n(2y)^{n}} {x-y^{n}}$.

For the proof of (ii), suppose y ⁿ > x. Letting $z = y -\frac{1} {k}$, with $k \in \mathbb{N}$, then, as in the previous paragraph anterior, we have y ⁿ > z ⁿ. We claim that z ⁿ > x for all sufficiently large k. To this end, arguing as above we have

$$\displaystyle{ \begin{array}{rl} z^{n}&\, = \left (y -\frac{1} {k}\right )^{n} = y^{n} +\sum _{ j=1}^{n}(-1)^{j}{n\choose j} \frac{1} {k^{j}}y^{n-j} \\ &\, > y^{n} -\sum _{j=1}^{n}{n\choose j} \frac{1} {k^{j}}y^{n-j} > y^{n} -\frac{n(2y)^{n}} {k}; \end{array} }$$

hence, z ⁿ > x whenever $y^{n} -\frac{n(2y)^{n}} {k} > x$, i.e., $k > \frac{n(2y)^{n}} {y^{n}-x}$. □ 

For another, conceptually simpler, proof of the above theorem, we refer the reader to Example 8.36.

Problems: Section 7.1

• 1. * Yet in respect to roots of positive real numbers, do the following items:

  • (a) If 0 < x < 1 and n > 1 is natural, show that the real number $\frac{1} {\root{n}\of{1/x}}$ (whose existence is guaranteed by Theorem 7.9, since $\frac{1} {x} > 1$) is the n-th root of x.

  • (b) Given a, b > 0 and m, n > 1 naturals, show that $\root{n}\of{ab} = \root{n}\of{a}\root{n}\of{b}$, $\root{n}\of{\frac{a} {b}} = \frac{\root{n}\of{a}} {\root{n}\of{b}}$ and $\root{mn}\of{a} = \root{m}\of{\root{n}\of{a}}$.

  • (c) Given 0 < a < b and n > 1 natural, show that $\root{n}\of{a} < \root{n}\of{b}$.

  • (d) Given a > 0 and m > n > 1 naturals, show that $\root{m}\of{a} > \root{n}\of{a}$ if a > 1, and $\root{m}\of{a} < \root{n}\of{a}$ if 0 < a < 1.

  • (e) Let x < 0 be a real number and $n \in \mathbb{N}$ be odd. If $y = -\root{n}\of{-x}$, show that y ⁿ = x (the real number y is also called the n-th root of x) and extend the properties of items (b)–(d) to this case.

• 2. * Use the archimedian property of the set of natural numbers to prove the following items:

  • (a) If $a \in \mathbb{R}$ is such that $0 \leq a < \frac{1} {n}$ for every $n \in \mathbb{N}$, then a = 0.

  • (b) If $a,b,c \in \mathbb{R}$, with a > 0, then there exists $n \in \mathbb{N}$ for which an + b > c.

• 3. * Let a and b be given rationals, with a < b. Prove that:

  • (a) $a < \frac{a+b} {2} < b$ and $a < a + \frac{b-a} {\sqrt{2}} < b$.

  • (b) The interval (a, b) contains infinitely many rational numbers and infinitely many irrational numbers.

• 4. * The purpose of this problem is to generalize the result of the previous one, showing that between any two given real numbers there is always a rational number and an irrational number (thanks to these properties, we say that $\mathbb{Q}$ and $\mathbb{R}\setminus \mathbb{Q}$ are dense in $\mathbb{R}$). To this end, let a and b be given real numbers, with a < b.

  • (a) Show that it suffices to consider the case a ≥ 0.

  • (b) Prove that there exists $n \in \mathbb{N}$ such that $0 < \frac{1} {n} < b - a$ and $0 < \frac{\sqrt{2}} {n} < b - a$.

  • (c) Letting a ≥ 0 e $n \in \mathbb{N}$ be chosen as in (b), show that one of the numbers $\frac{1} {n}, \frac{2} {n}, \frac{3} {n},\ldots$ and one of the numbers $\frac{\sqrt{2}} {n}, \frac{2\sqrt{2}} {n}, \frac{3\sqrt{2}} {n},\ldots$ belong to the interval (a, b).

• 5. * A rational number r ∈ [0, 1] is said to be dyadic if there exist $k,n \in \mathbb{Z}$ such that 0 ≤ n ≤ 2^k and $r = \frac{n} {2^{k}}$. Prove that the set of dyadic rationals is dense in [0, 1], i.e., that for every a ∈ [0, 1] and ε > 0, there exists a dyadic rational in the interval (a −ε, a +ε).

• 6. * If $Y \subset \mathbb{R}$ is nonempty and bounded below, prove that it has a greatest lower bound.

• 7. Let $X =\{ x \in \mathbb{Q};\,0 < x < 1\}$ and $Y =\{ y \in \mathbb{R}\setminus \mathbb{Q};\,0 < y < 1\}$. Prove that

  $$\displaystyle{\inf (X) =\inf (Y ) = 0 \text{and} \sup (X) =\sup (Y ) = 1.}$$

• 8. If $X =\{ \vert \sqrt{a} -\sqrt{b}\vert;\,a,b \in \mathbb{N}\,\,\text{and}\,\,a\neq b\}$, compute inf(X).

• 9. Let

  $$\displaystyle{ \begin{array}{rl} C_{1} & \, = (0,1) \\ C_{2} & \, = (0,1)\setminus (1/3,2/3) = (0,1/3] \cup [2/3,1) \\ C_{3} & \, = \left ((0,1/3]\setminus (1/9,2/9)\right ) \cup \left ([2/3,1)\setminus (7/9,8/9\right ) \\ &\, = (0,1/9] \cup [2/9,1/3] \cup [2/3,7/9] \cup [8/9,1).\end{array} }$$

  More generally, for each $n \in \mathbb{N}$, obtain C _n+1 from C _n , by erasing the open middle third of each of the intervals that form C _n . If $C =\bigcup _{n\geq 1}C_{n}$, show that inf(C) = 0 and sup(C) = 1.

• 10. * Let $Y \subset \mathbb{R}$ be a nonempty, bounded below set, with m = infY. If $n \in \mathbb{N}$, prove that there exists y ∈ Y such that

  $$\displaystyle{m \leq y < m + \frac{1} {n}.}$$

• 11. Let $X,Y \subset \mathbb{R}$ be nonempty sets, such that X is bounded above, Y is bounded below and sup(X) = inf(Y ) = α. If α ∉ X ∪ Y, prove that there exist elements x _n  ∈ X and y _n  ∈ Y for which $y_{n} - x_{n} < \frac{1} {n}$, for every $n \in \mathbb{N}$.

• 12. * Let $X \subset \mathbb{R}$ be a nonempty, bounded above set. Given $c \in \mathbb{R}$, let cX = { cx; x ∈ X}. Prove that:

  • (a) If c > 0, then sup(cX) = csup(X).

  • (b) If c < 0, then cX is bounded below and inf(cX) = csup(X).

  Then, if X is bounded below, establish properties analogous to the ones listed above, relating infX to the sup or the inf of cX, according to whether c < 0 or c > 0.

• 13. * Let $X,Y \subset \mathbb{R}$ be nonempty sets and X + Y = { x + y; x ∈ X and y ∈ Y }.

  • (a) If X and Y are bounded above, prove that X + Y is bounded above and sup(X + Y ) = sup(X) + sup(Y ).

  • (b) If X and Y are bounded below, prove that X + Y is bounded below and inf(X + Y ) = inf(X) + inf(Y ).

• 14. * Let X, Y ⊂ [0, +∞) be nonempty, bounded above sets. If XY = { xy; x ∈ X and y ∈ Y }, prove that XY is bounded above and such that sup(XY ) = sup(X) ⋅ sup(Y ) and inf(XY ) = inf(X) ⋅ inf(Y ).

• 15. (Hungary) Let (R _n ) _n ≥ 1 be a sequence of pairwise distinct rectangles in the cartesian plane, each of which having all vertices with integer coordinates and two sides along the axes. Prove that one can find two of them such that one contains the other.

• 16. (IMO) Let $f,g: \mathbb{R} \rightarrow \mathbb{R}$ be functions satisfying, for all $x,y \in \mathbb{R}$, the relation

  $$\displaystyle{f(x + y) + f(x - y) = 2f(x)g(y).}$$

  If f is not identically zero and | f(x) | ≤ 1 for every $x \in \mathbb{R}$, prove that | g(x) | ≤ 1, for every $x \in \mathbb{R}$.

7.2 Limits of Sequences

Given a sequence (a _n ) _n ≥ 1 of real numbers, we are interested in recognizing whether or not their terms are approaching a certain real number l, as n increases. For instance, if $a_{n} = \frac{1} {n}$, it is reasonable to say that the numbers a _n become closer and closer to 0 as n increases, since the result of the division of 1 by n is increasingly smaller as n increases. This naive point of view is formalized as follows.

Definition 7.10

A sequence (a _n ) _n ≥ 1 of real numbers converges to a real number l if, given an error ε > 0 for the value of l, there exists an index $n_{0} \in \mathbb{N}$ such that | a _n − l |  < ε for every n > n ₀.

Alternatively, if the sequence (a _n ) _n ≥ 1 converges to l, we say that it is convergent and that l is a limit of the sequence, which we denote by writing

$$\displaystyle{a_{n}\stackrel{n}{\longrightarrow }l \text{or} \lim _{n\rightarrow +\infty }a_{n} = l.}$$

Finally, a sequence which is not convergent is said to be divergent .

In general, if we diminish the error ε > 0 and the condition “ | a _n − l |  < ε for every n > n ₀” is to continue holding, then the natural number n ₀ in the definition of convergent sequence tends to increase. In other words, in general n ₀ depends on ε > 0. Anyhow, what is important to assure the convergence of the sequence (a _n ) _n ≥ 1 is that, for an arbitrarily given error ε > 0, we are capable of finding $n_{0} \in \mathbb{N}$ such that

$$\displaystyle{n > n_{0} \Rightarrow \vert a_{n} - l\vert <\epsilon.}$$

For the reader to get used to the important concept of convergent sequence, we collect below some elementary examples of convergent and divergent sequences.

Examples 7.11

  

• (a) If $a_{n} = \frac{1} {n}$, then $a_{n}\stackrel{n}{\longrightarrow }0$: indeed, for a given ε > 0, we have | a _n − 0 |  < ε provided $n > \frac{1} {\epsilon }$; thus, once we have chosen $n_{0} \in \mathbb{N}$ such that $n_{0} > \frac{1} {\epsilon }$, we will have | a _n − 0 |  < ε whenever n > n ₀.

• (b) If a _n  = (−1)ⁿ, then (a _n ) _n ≥ 1 is divergent: indeed, since the terms of the sequence are alternately equal to 1 and − 1, it is impossible for them to (collectively) become closer to a single real number l (formalize this intuition).

• (c) If $a_{n} = 1 + \frac{(-1)^{n}} {n}$, then $a_{n}\stackrel{n}{\longrightarrow }1$: this is so because $\vert a_{n} - 1\vert = \frac{1} {n}$, so that | a _n − 1 |  < ε for $n > \frac{1} {\epsilon }$.

• (d) If (a _n ) _n ≥ 1 is a constant sequence, with a _n  = c for every n ≥ 1, then a _n  → c.

Example 7.12

If a _n  = q ⁿ, with 0 <  | q |  < 1, then $a_{n}\stackrel{n}{\longrightarrow }0$.

Proof

Since $\frac{1} {\vert q\vert } > 1$, we can write $\frac{1} {\vert q\vert } = 1+\alpha$, with α > 0. Therefore, taking the first two terms in the binomial expansion formula, we get

$$\displaystyle{ \frac{1} {\vert q\vert ^{n}} = (1+\alpha )^{n} \geq 1 + n\alpha }$$

and, hence,

$$\displaystyle{\vert a_{n} - 0\vert = \vert q\vert ^{n} \leq \frac{1} {1 + n\alpha }.}$$

Thus, if we wish that | a _n − 0 |  < ε, it suffices to impose $\frac{1} {1+n\alpha } <\epsilon$ or, equivalently, $n > \frac{1} {\alpha } \left (\frac{1} {\epsilon } - 1\right )$. □ 

Example 7.13

The sequence (a _n ) _n ≥ 1, given for n ≥ 1 by $a_{n} = \sqrt{n + 1} -\sqrt{n}$, converges to 0.

Proof

Note that $a_{n} = \frac{1} {\sqrt{n+1}+\sqrt{n}}$. Thus, given ε > 0 and taking $n_{0} \in \mathbb{N}$ such that $n_{0} > \frac{1} {4\epsilon ^{2}}$, we have

$$\displaystyle{n > n_{0} \Rightarrow \sqrt{n + 1} + \sqrt{n} > \sqrt{n_{0 } + 1} + \sqrt{n_{0}} > 2\sqrt{n_{0}} > \frac{1} {\epsilon }.}$$

Therefore,

$$\displaystyle{n > n_{0} \Rightarrow \vert a_{n} - 0\vert = \frac{1} {\sqrt{n + 1} + \sqrt{n}} <\epsilon.}$$

 □ 

The notion of convergent sequence doesn’t make it clear whether the correspondent limit is unique. Yet in another way, in principle it could happen that a certain sequence converges to more than one limit. The coming result shows that this is not so.

Proposition 7.14

If the sequence (a _n ) _n≥1 converges, then its limit is unique.

Proof

Let l ₁ and l ₂ be distinct real numbers, and suppose that the given sequence simultaneously converges to l ₁ and l ₂. Toking $\epsilon = \frac{1} {2}\vert l_{1} - l_{2}\vert > 0$, the definition of convergence guarantees the existence of $n_{1},n_{2} \in \mathbb{N}$ such that

$$\displaystyle{n > n_{1} \Rightarrow \vert a_{n} - l_{1}\vert <\epsilon \ \ \text{and}\ \ n > n_{2} \Rightarrow \vert a_{n} - l_{2}\vert <\epsilon.}$$

Therefore, triangle inequality gives

$$\displaystyle{n >\max \{ n_{1},n_{2}\} \Rightarrow \vert l_{1} - l_{2}\vert \leq \vert a_{n} - l_{1}\vert + \vert a_{n} - l_{2}\vert < 2\epsilon = \vert l_{1} - l_{2}\vert,}$$

which is an absurd. □ 

Thanks to the previous result, from now on we speak of the limit of a convergent sequence. In this respect, the next proposition collects two basic, albeit very important, properties of limits of convergent sequences. In order to state it properly, we define a subsequence of a sequence (a _n ) _n ≥ 1 as the restriction of the given sequence to an infinite subset $\mathbb{N}_{1} =\{ n_{1} < n_{2} < n_{3} < \cdots \,\}$ of $\mathbb{N}$; in this case, we denote it by $(a_{n_{k}})_{k\in \mathbb{N}}$. Since the function j ↦ n _j from $\mathbb{N}_{1}$ to $\mathbb{N}$ is a bijection, every subsequence can actually be seen as a sequence.

Proposition 7.15

Let (a _n ) _n≥1 be a convergent sequence, with lim_n→+∞ a _n = l. Then:

• (a) If a _n ≥ a ( resp. a _n ≤ a ) , for every n ≥ 1, then l ≥ a ( resp. l ≤ a ) .

• (b) Every subsequence $(a_{n_{k}})_{k\geq 1}$ of (a _n ) _n≥1 also converges to l.

Proof

  

• (a) Suppose that a _n  ≥ a for every n ≥ 1, and let’s show that l ≥ a (the other case is completely analogous). By contradiction, if l < a, take ε = a − l > 0. The definition of convergence guarantees the existence of an index $n_{0} \in \mathbb{N}$ such that n > n ₀ ⇒ | a _n − l |  < ε; in particular, given n > n ₀, we have

  $$\displaystyle{a_{n} < l+\epsilon = l + (a - l) = a,}$$

  which is an absurd.

• (b) Let ε > 0 be given. Since $a_{n}\stackrel{n}{\longrightarrow }l$, there exists a natural number n ₀ such that | a _n − l |  < ε for n > n ₀. Since n ₁ < n ₂ < n ₃ < ⋯ , there exists an index n _i such that n _j  > n ₀ for j ≥ i; hence, for all such j, we have $\vert a_{n_{j}} - l\vert <\epsilon$, which is the same as saying that $a_{n_{k}}\stackrel{k}{\longrightarrow }l$. □ 

In words, item (b) of the previous proposition says that, if the terms of a certain sequence come closer and closer to l as their indices increase, then the same is true for the terms of every subsequence of the given sequence. Item (b) of the previous proposition also has the following immediate corollary, which gives us a sufficient (and quite useful) condition for the divergence of a sequence. sequência.

Corollary 7.16

If two subsequences of a given sequence converge to distinct limits, then the original sequence is divergent.

Up to now, except for some very simple examples we haven’t seen how one could find out the limit of a convergent sequence. In order to remedy this situation, we need to understand how to perform simple arithmetic operations with convergent sequences. We turn to this next, starting with an auxiliary result which is important in its own.

We say that a sequence (a _n ) _n ≥ 1 is bounded (resp. bounded from above, bounded from below ) if the set {a ₁, a ₂, …} is bounded (resp. bounded from above, bounded from below), in the sense of the previous section.

Lemma 7.17

Every convergent sequence is bounded.

Proof

If (a _n ) _n ≥ 1 is a convergent sequence with limit l, then there exists $n_{0} \in \mathbb{N}$ such that

$$\displaystyle{n > n_{0} \Rightarrow \vert a_{n} - l\vert < 1.}$$

This, together with the triangle inequality, gives

$$\displaystyle{n > n_{0} \Rightarrow \vert a_{n}\vert \leq \vert a_{n} - l\vert + \vert l\vert < 1 + \vert l\vert.}$$

Finally, letting $L =\max \{ 1 + \vert a\vert,\vert a_{1}\vert,\vert a_{2}\vert,\ldots,\vert a_{n_{0}-1}\vert \}$, we get | a _n  |  < L for every $n \in \mathbb{N}$, so that the sequence is bounded. □ 

Proposition 7.18

Let (a _n ) _n≥1 and (b _n ) _n≥1 be convergent sequences, and c be any real number.

• (a) If $a_{n}\stackrel{n}{\longrightarrow }a$ , then $ca_{n}\stackrel{n}{\longrightarrow }ca$ .

• (b) If $a_{n}\stackrel{n}{\longrightarrow }a$ and $b_{n}\stackrel{n}{\longrightarrow }b$ , then $a_{n} \pm b_{n}\stackrel{n}{\longrightarrow }a \pm b$ and $a_{n}b_{n}\stackrel{n}{\longrightarrow }ab$ .

• (c) If $a_{n}\stackrel{n}{\longrightarrow }0$ and (b _n ) _n≥1 is bounded, then $a_{n}b_{n}\stackrel{n}{\longrightarrow }0$ .

• (d) If $a_{n}\stackrel{n}{\longrightarrow }a$ and $b_{n}\stackrel{n}{\longrightarrow }b$ , with b,b _n ≠ 0 for every n ≥ 1, then $\frac{a_{n}} {b_{n}} \stackrel{n}{\longrightarrow }\frac{a} {b}$ .

Proof

  

• (a) If c = 0, then ca _n  = ca = 0, and there is nothing to do. Suppose, then, that c ≠ 0, and let ε > 0 be given. Since $a_{n}\stackrel{n}{\longrightarrow }a$, there exists $n_{0} \in \mathbb{N}$ such that $n > n_{0} \Rightarrow \vert a_{n} - a\vert < \frac{\epsilon } {\vert c\vert }$. Hence,

  $$\displaystyle{n > n_{0} \Rightarrow \vert ca_{n} - ca\vert = \vert c\vert \vert a_{n} - a\vert < \vert c\vert \cdot \frac{\epsilon } {\vert c\vert } =\epsilon.}$$

• (b) For the first part, let’s prove that $a_{n} + b_{n}\stackrel{n}{\longrightarrow }a + b$ (to prove that a _n − b _n  → a − b is completely analogous). Given ε > 0, the convergences $a_{n}\stackrel{n}{\longrightarrow }a$ and $b_{n}\stackrel{n}{\longrightarrow }b$ assure the existence of $n_{1},n_{2} \in \mathbb{N}$ such that

  $$\displaystyle{n > n_{1} \Rightarrow \vert a_{n} - a\vert < \frac{\epsilon } {2} \text{and} n > n_{2} \Rightarrow \vert b_{n} - b\vert < \frac{\epsilon } {2}.}$$

  Therefore, taking n > max{n ₁, n ₂}, we get

  $$\displaystyle{\vert (a_{n} + b_{n}) - (a + b)\vert \leq \vert a_{n} - a\vert + \vert b_{n} - b\vert < \frac{\epsilon } {2} + \frac{\epsilon } {2} =\epsilon.}$$

  For the second part, let L > 0 be such that | b _n  |  < L for every $n \in \mathbb{N}$. Given ε > 0, take $n_{0} \in \mathbb{N}$ for which

  $$\displaystyle{n > n_{0} \Rightarrow \vert a_{n} - a\vert < \frac{\epsilon } {2L} \text{and} \vert b_{n} - b\vert < \frac{\epsilon } {2\vert a\vert + 1}.}$$

  Then

  $$\displaystyle{ \begin{array}{rl} \vert a_{n}b_{n} - ab\vert &\, = \vert a_{n}b_{n} - ab_{n} + ab_{n} - ab\vert \leq \vert a_{n} - a\vert \vert b_{n}\vert + \vert a\vert \vert b_{n} - b\vert \\ &\, < \frac{\epsilon } {2L} \cdot L + \vert a\vert \cdot \frac{\epsilon } {2\vert a\vert +1} < \frac{\epsilon } {2} + \frac{\epsilon } {2} =\epsilon.\end{array} }$$

• (c) Let L > 0 be such that | b _n  |  < L for every n ≥ 1. Given ε > 0, let’s take $n_{0} \in \mathbb{N}$ such that

  $$\displaystyle{n > n_{0} \Rightarrow \vert a_{n}\vert < \frac{\epsilon } {L}.}$$

  Then,

  $$\displaystyle{n > n_{0} \Rightarrow \vert a_{n}b_{n} - 0\vert = \vert a_{n}\vert \vert b_{n}\vert < \frac{\epsilon } {L} \cdot L =\epsilon.}$$

• (d) By the second part of item (b), it suffices to show that $\frac{1} {b_{n}}\stackrel{n}{\longrightarrow }\frac{1} {b}$. To this end, start by observing that

  $$\displaystyle{\left \vert \frac{1} {b_{n}} -\frac{1} {b}\right \vert = \frac{1} {\vert b\vert }\cdot \frac{\vert b_{n} - b\vert } {\vert b_{n}\vert } \leq \frac{1} {\vert b\vert }\cdot \frac{\vert b_{n} - b\vert } {\vert b\vert -\vert b_{n} - b\vert },}$$

  where we used the triangle inequality in the last passage above. Now, given ε > 0, choose $n_{0} \in \mathbb{N}$ such that

  $$\displaystyle{n > n_{0} \Rightarrow \vert b_{n} - b\vert < \frac{\epsilon } {2}, \frac{\vert b\vert } {2}.}$$

  Then, for n > n ₀, we have

  $$\displaystyle{\left \vert \frac{1} {b_{n}} -\frac{1} {b}\right \vert \leq \frac{1} {\vert b\vert }\cdot \frac{\vert b_{n} - b\vert } {\vert b\vert -\vert b\vert /2} = 2\vert b_{n} - b\vert <\epsilon.}$$

   □ 

Example 7.19

Let a be a positive real number. If (a _n ) _n ≥ 1 is given by $a_{n} = \root{n}\of{a}$ for every n ≥ 1, then $a_{n}\stackrel{n}{\longrightarrow }1$.

Proof

If a > 1, then a _n  > 1. Write a _n  = 1 + b _n , so that b _n  > 0. Since

$$\displaystyle{a = a_{n}^{n} = (1 + b_{ n})^{n} \geq 1 +{ n\choose 1}b_{ n} = 1 + nb_{n},}$$

we get $0 < b_{n} < \frac{a-1} {n}$. Hence, the squeezing principle (cf. Problem 6) guarantees that $b_{n}\stackrel{n}{\longrightarrow }0$, and item (b) of Proposition 7.18 gives $a_{n} = 1 + b_{n}\stackrel{n}{\longrightarrow }1$.

If 0 < a < 1, let $a_{n}^{{\prime}} = \frac{1} {a_{n}} = \root{n}\of{\frac{1} {a}}$, so that $a_{n}^{{\prime}}\stackrel{n}{\longrightarrow }1$ by the first part. Then, item (d) of Proposition 7.18 gives that $a_{n}\stackrel{n}{\longrightarrow }1$. □ 

Example 7.20

The sequence (a _n ) _n ≥ 1, given by $a_{n} = \root{n}\of{n}$ for every n ≥ 1, converges to 1.

Proof

As in the previous example, write a _n  = 1 + b _n for n ≥ 2. Since b _n  > 0, we have

$$\displaystyle{n = a_{n}^{n} = (1 + b_{ n})^{n} \geq 1 +{ n\choose 1}b_{ n} +{ n\choose 2}b_{n}^{2} > \frac{n(n - 1)} {2} \cdot b_{n}^{2},}$$

so that

$$\displaystyle{0 < b_{n}^{2} < \frac{2} {n - 1}.}$$

Hence, once more from the squeezing principle, the above inequality gives $b_{n}\stackrel{n}{\longrightarrow }0$ and, thus, $a_{n} = 1 + b_{n}\stackrel{n}{\longrightarrow }1$. □ 

For what comes next, recall that a sequence (a _n ) _n ≥ 1 of real numbers is just a function $f: \mathbb{N} \rightarrow \mathbb{R}$, for which we write a _n  = f(n). Hence, it is natural to say that (a _n ) _n ≥ 1 is monotonic increasing (resp. decreasing, nondecreasing, nonincreasing) provided a _n  < a _n+1 (resp. a _n  > a _n+1, a _n  ≤ a _n+1, a _n  ≥ a _n+1), for every n ≥ 1.

The most important result on limits of sequences is the theorem below, which is known in the mathematical literature as Bolzano-Weierstrass theorem .³

Theorem 7.21 (Bolzano-Weierstrass)

Every monotonic bounded sequence is convergent.

Proof

Suppose that (a _n ) _n ≥ 1 is a nondecreasing bounded sequence (the other cases can be dealt with similarly), i.e., that

$$\displaystyle{a_{1} \leq a_{2} \leq a_{3} \leq \cdots < M,}$$

for some M > 0. Then, M is an upper bound for the set A = { a ₁, a ₂, a ₃, …}, so that A has a sup, say supA = l. We claim that $a_{n}\stackrel{n}{\longrightarrow }l$. Indeed, let ε > 0 be given; since l −ε is no longer an upper bound for A, some element of it is greater than l −ε, say, $a_{n_{0}} > l-\epsilon$. Therefore, since $a_{n_{0}} \leq a_{n_{0}+1} \leq a_{n_{0}+2} \leq \cdots$, we conclude that a _n  > l −ε for every n ≥ n ₀. Thus, for n ≥ n ₀, we have

$$\displaystyle{l-\epsilon < a_{n} \leq l < l+\epsilon,}$$

as we wished to show. □ 

The previous theorem, together with the definition of convergence, assures that if a bounded sequence is monotonic from a certain term on, then it will be convergent. We explore this comment by revisiting the last two examples.

Example 7.22

Let a be a positive real number. If (a _n ) _n ≥ 1 is given by $a_{n} = \root{n}\of{a}$ for every n ≥ 1, then $a_{n}\stackrel{n}{\longrightarrow }1$.

Proof

Assume a > 1 (the case 0 < a < 1 can be dealt with as in the Example 7.19). Then, a ₁ > a ₂ > a ₃ > ⋯ > 1, and the Bolzano-Weierstrass theorem guarantees the existence of l = lim _{n → +∞} a _n  ≥ 1. Item (a) of Proposition 7.15 gives l ≥ 1, and item (b) guarantees that every subsequence of (a _n ) _n ≥ 1 also converges to l. Therefore, $a_{k(k+1)}\stackrel{k}{\longrightarrow }l$. Now, since $a_{k(k+1)} = \frac{\root{k}\of{a}} {\root{k+1}\of{a}}$, it follows from item (d) of Proposition 7.18 that

$$\displaystyle{a_{k(k+1)} = \frac{\root{k}\of{a}} {\root{k + 1}\of{a}}\stackrel{k}{\longrightarrow }\frac{l} {l} = 1.}$$

Therefore, l = 1. □ 

Example 7.23

The sequence (a _n ) _n ≥ 1, given by $a_{n} = \root{n}\of{n}$ for every n ≥ 1, converges to 1.

Proof

The initial terms of the sequence are $\sqrt{2}$, $\root{3}\of{3}$, $\root{4}\of{4}$, …, and it is easy to directly show that $\sqrt{2} < \root{3}\of{3}$ and $\root{3}\of{3} > \root{4}\of{4} > \root{5}\of{5}$. Since 2ⁿ ≥ n ² for n ≥ 4 (by induction, for instance), we get a ₂ ≥ a _n  > 1 for n ≥ 4, so that the sequence is bounded; hence, if we show that it is indeed decreasing from the third term on, its convergence will follow from Bolzano-Weierstrass theorem, with limit l ≥ 1.

For what is left to do, for an integer n > 2 we have

$$\displaystyle{\root{n}\of{n} > \root{n + 1}\of{n + 1} \Leftrightarrow n^{n+1} > (n + 1)^{n} \Leftrightarrow n > \left (1 + \frac{1} {n}\right )^{\!n}.}$$

Let’s prove the last inequality above. For n = 3, it’s immediate to check it numerically; for n > 3, it suffices to show that $\left (1 + \frac{1} {n}\right )^{n} < 3$. To this end, notice that

$$\displaystyle{\left (1 + \frac{1} {n}\right )^{n} = 1 +{ n\choose 1} \frac{1} {n} +{ n\choose 2} \frac{1} {n^{2}} + \cdots +{ n\choose n} \frac{1} {n^{n}}}$$

and

$$\displaystyle{{n\choose k} \frac{1} {n^{k}} = \frac{n!} {k!(n - k)!n^{k}} = \frac{1} {k!} \cdot \frac{n(n - 1)\ldots (n - k + 1)} {n^{k}} < \frac{1} {k!} \leq \frac{1} {2^{k-1}}.}$$

Therefore,

$$\displaystyle{ \begin{array}{rl} \left (1 + \frac{1} {n}\right )^{n}&\, = 1 +{ n\choose 1} \frac{1} {n} +{ n\choose 2} \frac{1} {n^{2}} + \cdots +{ n\choose n} \frac{1} {n^{n}} \\ & \, < 1 + 1 + \frac{1} {2} + \frac{1} {2^{2}} + \cdots + \frac{1} {2^{n-1}} \\ & \, = 3 - \frac{1} {2^{n-1}} < 3. \end{array} }$$

In order to finish, we need to show that l = 1. To this end, first observe that the subsequence $a_{2k} = \root{2k}\of{2k}$ also converges to l. On the other hand, $\root{2k}\of{2k} = \root{2k}\of{2} \cdot \sqrt{\root{k}\of{k}}$, with $\root{2k}\of{2}\stackrel{k}{\longrightarrow }1$. Now, it follows from Problem 2 that $\sqrt{\root{k}\of{k}}\stackrel{k}{\longrightarrow }\sqrt{l}$. Therefore, by applying item (b) of Proposition 7.18, we get

$$\displaystyle{l =\lim _{k\rightarrow +\infty }\root{2k}\of{2k} =\lim _{k\rightarrow +\infty }\root{2k}\of{2} \cdot \lim _{k\rightarrow +\infty }\sqrt{\root{k}\of{k}} = \sqrt{l},}$$

 □ 

Sometimes, we have to show that a given sequence has at least a convergent subsequence (even if, as a whole, the sequence does not converge). In this sense, Theorem 7.25 below, also due to Weierstrass, provides a sufficient condition for the existence of such a subsequence. Before we state and prove it, we need to discuss an important auxiliary result, known as the lemma of nested intervals . In what follows if I = [a, b], with $a,b \in \mathbb{R}$, we shall let | I |  = b − a.

Lemma 7.24

For $n \in \mathbb{N}$ , let I _n = [a _n ,b _n ]. If I ₁ ⊃ I ₂ ⊃ I ₃ ⊃ … and lim_n→+∞ |I _n | = 0, then there exists a unique $l \in \mathbb{R}$ such that $\bigcap _{n\geq 1}I_{n} =\{ l\}$ .

Proof

First of all, note that the intersection of the intervals I _n , if not empty, has a single element; indeed, if there existed reals a < b in such an intersection, we would have $[a,b] \subset \bigcap _{n\geq 1}I_{n}$; in particular, [a, b] ⊂ I _n and, hence, | I _n  | ≥ b − a for every $n \in \mathbb{N}$, thus contradicting the fact that lim _{n → +∞}  | I _n  |  = 0.

Secondly, the inclusions I ₁ ⊃ I ₂ ⊃ I ₃ ⊃ … give a ₁ ≤ a ₂ ≤ a ₃ ≤ ⋯  ≤ ⋯ ≤ b ₃ ≤ b ₂ ≤ b ₁, and the Bolzano-Weierstrass theorem assures the existence of l = lim _{n → +∞} a _n . We claim that $l \in \bigcap _{n\geq 1}I_{n}$.

Since $l =\sup \{ a_{n};\,n \in \mathbb{N}\}$, it follows that a _n  ≤ l for every $n \in \mathbb{N}$. On the other hand, for a fixed $m \in \mathbb{N}$, we have a _n  ≤ b _m for every $n \in \mathbb{N}$, and item (a) of Proposition 7.15 gives l = lim _{n → +∞} a _n  ≤ b _m . However, since m was chosen arbitrarily, we have l ≤ b _m for every $m \in \mathbb{N}$. It thus follows that l ∈ [a _m , b _m ] = I _m for every $m \in \mathbb{N}$, as we wished to show. □ 

We are now in position to prove Weierstrass theorem.

Theorem 7.25 (Weierstrass)

Every bounded sequence admits a convergent subsequence.

Proof

Let (a _n ) _n ≥ 1 be a given bounded sequence, and I ₀ = [a, b] be a closed and bounded interval containing all of its terms. One (possibly both) of the intervals $\left [a, \frac{a+b} {2} \right ]$ and $\left [\frac{a+b} {2},b\right ]$, call it I ₁, also contains infinitely many terms of the sequence. Do the same with I ₁, obtaining a closed and bounded interval I ₂ ⊂ I ₁ such that $\vert I_{2}\vert = \frac{1} {2}\vert I_{1}\vert$ and I ₂ contains infinitely many terms of the sequence (a _n ) _n ≥ 1. Proceeding inductively, we construct a nested sequence I ₁ ⊃ I ₂ ⊃ I ₃ ⊃ ⋯ of closed and bounded intervals, such that $\vert I_{k+1}\vert = \frac{1} {2}\vert I_{k}\vert$ and I _k contains infinitely many terms of the sequence (a _n ) _n ≥ 1, for every k ≥ 1. Therefore, by the lemma of nested intervals, there exists $c \in \mathbb{R}$ such that $\bigcap _{k\geq 1}I_{k} =\{ c\}$.

Now, choose $n_{1} \in \mathbb{N}$ such that $a_{n_{1}} \in I_{1}$; then, after having chosen $n_{j} \in \mathbb{N}$ such that $a_{n_{j}} \in I_{j}$, choose $n_{j+1} \in \mathbb{N}$ such that n _j+1 > n _j and $a_{n_{j+1}} \in I_{j+1}$ (this is possible by the way the I _j ’s were defined). This way, we inductively construct a subsequence $(a_{n_{k}})_{k\geq 1}$ of (a _n ) _n ≥ 1, such that $a_{n_{k}} \in I_{k}$ for every k ≥ 1. Since $\vert I_{k}\vert = \frac{1} {2^{k}}\vert I_{0}\vert$ and $a_{n_{k}},c \in I_{k}$, we conclude that $\vert a_{n_{k}} - c\vert \leq \frac{1} {2^{k}}\vert I_{0}\vert$, for every k ≥ 1; since $\frac{1} {2^{k}}\vert I_{0}\vert \stackrel{k}{\longrightarrow }0$, we conclude that $a_{n_{k}}\stackrel{k}{\longrightarrow }c$. □ 

The concept of convergent sequence gives a precise meaning to the geometric intuition that the terms of the given sequence come closer and closer to a certain real number (the limit of the sequence), as long as their indices increase. However, it is also reasonable to expect that, if the terms of a given sequence come close together, then the sequence should also converge. This is indeed the case and, in order to establish it, we start with the following

Definition 7.26

A sequence (a _n ) _n ≥ 1 is said to be a Cauchy sequence if, for every ε > 0, there exists $n_{0} \in \mathbb{N}$ such that

$$\displaystyle{m,n > n_{0} \Rightarrow \vert a_{m} - a_{n}\vert <\epsilon.}$$

The fundamental result concerning Cauchy sequences is the content of the following

Theorem 7.27

A sequence (a _n ) _n≥1 is convergent if and only if is Cauchy.

Proof

Let (a _n ) _n ≥ 1 be a convergent sequence, with limit l. Given ε > 0, the definition of convergence guarantees the existence of $n_{0} \in \mathbb{N}$ such that

$$\displaystyle{n > n_{0} \Rightarrow \vert a_{n} - l\vert < \frac{\epsilon } {2}.}$$

Hence, given naturals m, n > n ₀, the triangle inequality gives

$$\displaystyle{\vert a_{m} - a_{n}\vert \leq \vert a_{m} - l\vert + \vert a_{n} - l\vert < \frac{\epsilon } {2} + \frac{\epsilon } {2} =\epsilon,}$$

and the sequence is Cauchy.

Conversely, let (a _n ) _n ≥ 1 be a Cauchy sequence. Then, there exists $n_{0} \in \mathbb{N}$ such that | a _m − a _n  |  < 1 for m, n > n ₀. In particular, $\vert a_{m} - a_{n_{0}+1}\vert < 1$ for every m > n ₀, and the sequence has all of its terms contained in the set

$$\displaystyle{\{a_{1},a_{2},\ldots,a_{n_{0}}\} \cup (a_{n_{0}+1} - 1,a_{n_{0}+1} + 1),}$$

so that it is bounded. Hence, by the theorem of Bolzano-Weierstrass, the sequence (a _n ) _n ≥ 1 has a convergent subsequence, say, $a_{n_{k}}\stackrel{k}{\longrightarrow }l$. Let us prove that, actually, $a_{n}\stackrel{n}{\longrightarrow }l$.

Given ε > 0, there exists $N_{0} \in \mathbb{N}$ such that

$$\displaystyle{n_{k} > N_{0} \Rightarrow \vert a_{n_{k}} - l\vert < \frac{\epsilon } {2}.}$$

On the other hand, since the sequence is Cauchy, there exists $N_{1} \in \mathbb{N}$ such that

$$\displaystyle{m,n > N_{1} \Rightarrow \vert a_{m} - a_{n}\vert < \frac{\epsilon } {2}.}$$

Letting M = max{N ₁, N ₂} and fixing n _k  > M, we have, for n > M and with the aid of the triangle inequality,

$$\displaystyle{\vert a_{n} - l\vert \leq \vert a_{n} - a_{n_{k}}\vert + \vert a_{n_{k}} - l\vert < \frac{\epsilon } {2} + \frac{\epsilon } {2} =\epsilon,}$$

as we wished to show. □ 

The coming example collects an interesting application of the above result.

Example 7.28

Let (a _n ) _n ≥ 1 be a sequence of real numbers such that

$$\displaystyle{\vert a_{n+2} - a_{n+1}\vert \leq c\vert a_{n+1} - a_{n}\vert }$$

for every $n \in \mathbb{N}$, where 0 < c < 1 is a real constant. Show that this sequence is convergent.

Proof

By Theorem 7.27, it suffices to show that (a _n ) _n ≥ 1 is a Cauchy sequence. To this end, iterating the inequality in the statement we get, for every $k \in \mathbb{N}$,

$$\displaystyle{\vert a_{k+1} - a_{k}\vert \leq c^{k-1}\vert a_{ 2} - a_{1}\vert.}$$

Let n and p be given natural numbers. The above inequality, together with the triangle inequality, gives

$$\displaystyle{ \begin{array}{rl} \vert a_{n+p} - a_{n}\vert &\,\leq \sum _{k=n}^{n+p-1}\vert a_{k+1} - a_{k}\vert \leq \sum _{k=n}^{n+p-1}c^{k-1}\vert a_{2} - a_{1}\vert \\ &\, = \left (\frac{c^{n-1}-c^{n+p}} {1-c} \right )\vert a_{2} - a_{1}\vert \\ &\, < \frac{1} {1-c} \cdot \vert a_{2} - a_{1}\vert c^{n-1}. \end{array} }$$

We now note that, by Example 7.12, the last expression above tends to 0 when n → +∞. Therefore, given ε > 0, there exists $n_{0} \in \mathbb{N}$ such that $\frac{1} {1-c} \cdot \vert a_{2} - a_{1}\vert c^{n-1} <\epsilon$ for every n > n ₀. Hence, for n > n ₀ and $p \in \mathbb{N}$, we get | a _n+p − a _n  |  < ε, so that (a _n ) _n ≥ 1 is, indeed, a Cauchy sequence. □ 

Problems: Section 7.2

• 1. * Let (a _n ) _n ≥ 1 and (b _n ) _n ≥ 1 be convergent sequences of real numbers, with lim _{n → +∞} a _n  = a and lim _{n → +∞} b _n  = b. Generalize item (a) of Proposition 7.15, showing that if a _n  ≤ b _n for every n ≥ 1, then a ≤ b.

• 2. * Let (a _n ) _n ≥ 1 be a sequence of positive real numbers converging to a > 0. Show that $\sqrt{a_{n}}\stackrel{n}{\longrightarrow }\sqrt{a}$.

• 3. * Given $a \in \mathbb{R}$ such that | a |  > 1, show that $\frac{a^{n}} {n!} \rightarrow 0$ when n → +∞.

• 4. Generalize the result of Example 7.12, showing that, given $k \in \mathbb{N}$ and $a \in \mathbb{R}$, with | a |  > 1, we have $\frac{n^{k}} {a^{n}}\stackrel{n}{\longrightarrow }0$.

• 5. Let (a _n ) _n ≥ 1 and (b _n ) _n ≥ 1 be sequences of real numbers and, for each $n \in \mathbb{N}$, let t _n  ∈ [0, 1] be given. Denote by (c _n ) _n ≥ 1 the sequence defined by

  $$\displaystyle{c_{n} = (1 - t_{n})a_{n} + t_{n}b_{n},}$$

  for every $n \in \mathbb{N}$. If $a_{n},b_{n}\stackrel{n}{\longrightarrow }l$, show that $c_{n}\stackrel{n}{\longrightarrow }l$.

• 6. * Prove the squeezing theorem : let (a _n ) _n ≥ 1, (b _n ) _n ≥ 1 and (c _n ) _n ≥ 1 be sequences of real numbers such that a _n  ≤ b _n  ≤ c _n , for every $n \in \mathbb{N}$. If $a_{n},c_{n}\stackrel{n}{\longrightarrow }l$, for some $l \in \mathbb{R}$, show that $b_{n}\stackrel{n}{\longrightarrow }l$.

• 7. Compute the following limits:

  • (a) $\lim _{n\rightarrow +\infty }\frac{n\sqrt{n}} {n^{2}+1}$.

  • (b) $\lim _{n\rightarrow +\infty }(\sqrt{n^{2 } + an + b} - n)$, with $a,b \in \mathbb{R}$.

  • (c) $\lim _{n\rightarrow +\infty }\root{n}\of{1 + q^{n}}$, where 0 < q < 1 is a real number.

  • (d) $\lim _{n\rightarrow +\infty }\root{n}\of{a^{n} + b^{n}}$, with a and b positive reals such that a > b.

• 8. * This problem extends the concept of limit of sequences to consider infinite limits. We say that a sequence (a _n ) _n ≥ 1 of real numbers converges to + ∞ (resp. −∞) if, given M > 0, there exists $n_{0} \in \mathbb{N}$ such that n > n ₀ ⇒ a _n  > M (resp. a _n  < −M). In this case, we denote lim _{n → +∞} a _n  = +∞ (resp. lim _{n → +∞} a _n  = −∞), or simply $a_{n}\stackrel{n}{\longrightarrow } + \infty$ (resp. $a_{n}\stackrel{n}{\longrightarrow } -\infty$). With respect to this concept, and given sequences (a _n ) _n ≥ 1 and (b _n ) _n ≥ 1 of real numbers, do the following items:

  • (a) If $a_{n}\stackrel{n}{\longrightarrow } \pm \infty$ and (b _n ) _n ≥ 1 is bounded, then $a_{n} + b_{n}\stackrel{n}{\longrightarrow } \pm \infty$.

  • (b) If $a_{n}\stackrel{n}{\longrightarrow } \pm \infty$ and b _n  ≥ c > 0 (resp. b _n  ≤ c < 0) for every n ≥ 1, then $a_{n}b_{n}\stackrel{n}{\longrightarrow } \pm \infty$ (resp. $a_{n}b_{n}\stackrel{n}{\longrightarrow } \mp \infty$).

  • (c) If $b_{n}\stackrel{n}{\longrightarrow } + \infty$ and there exists c > 0 such that a _n  ≥ cb _n (resp. a _n  ≤ −cb _n ) for every n ≥ 1, then $a_{n}\stackrel{n}{\longrightarrow } \pm \infty$.

• 9. Let q be a real number and (a _n ) _n ≥ 1 be the sequence defined by a _n  = q ⁿ. If q > 1, show that $a_{n}\stackrel{n}{\longrightarrow } + \infty$. If q < −1, show that (a _n ) _n ≥ 1 does not converge to either + ∞ or −∞.

• 10. * Given positive reals a and q, with q < 1, show that $a^{n}q^{2^{n} }\stackrel{n}{\longrightarrow }0$.

• 11. (IMO shortlist) Let (a _n ) _n ≥ 1 be a sequence of real numbers such that, for every $m,n \in \mathbb{N}$, we have

  $$\displaystyle{\vert a_{m} - a_{n}\vert \leq \frac{2mn} {m^{2} + n^{2}}.}$$

  Show that the sequence is constant.

• 12. Let (a _n ) _n ≥ 1 be the sequence defined by a ₁ = 1, $a_{2} = \sqrt{1 + 1}$, $a_{3} = \sqrt{1 + \sqrt{2}}$, $a_{4} = \sqrt{1 + \sqrt{1 + \sqrt{2}}}$, $a_{5} = \sqrt{1 + \sqrt{1 + \sqrt{1 + \sqrt{2}}}}$, …. Show that (a _n ) _n ≥ 1 is convergent and compute its limit.

• 13. (Austrian-Polish) Let (a _n ) _n ≥ 1 be a sequence of positive reals, such that

  $$\displaystyle{a_{k+2} = \sqrt{a_{k+1}} + \sqrt{a_{k}},}$$

  for every k ≥ 1. Prove that the sequence converges and compute its limit.

• 14. Let n > 1 be a given integer and t ₀, t ₁, …, t _n be given real numbers, such that t ₀ + t ₁ + ⋯ + t _n  = 0. Prove that the sequence (a _k ) _k ≥ 1, defined by

  $$\displaystyle{a_{k} = t_{0}\sqrt{k} + t_{1}\sqrt{k + 1} + \cdots + t_{n}\sqrt{k + n}}$$

  converges to 0.

• 15. (Romania) The sequence (x _n ) _n ≥ 1 is such that $\sqrt{x_{n+1 } + 2} \leq x_{n} \leq 2$, for every n ≥ 1. Find all possible values of x ₁₉₈₆.

• 16. (Leningrad) Let (a _n ) _n ≥ 1 be a sequence of real numbers such that

  $$\displaystyle{\vert a_{m} + a_{n} - a_{m+n}\vert \leq \frac{1} {m + n},}$$

  for all $m,n \in \mathbb{N}$. Prove that the sequence is an AP.

• 17. (Bulgaria) For each $n \in \mathbb{N}$, let

  $$\displaystyle{a_{n} = \frac{n + 1} {2^{n+1}} \left (\frac{2^{1}} {1} + \frac{2^{2}} {2} + \cdots + \frac{2^{n}} {n} \right ).}$$

  Prove that the sequence (a _n ) _n ≥ 2 is decreasing and convergent, and compute its limit.

• 18. (Romania) Let k be a fixed natural number and (a _n ) _n ≥ 1 be the sequence defined by

  $$\displaystyle{a_{n} = \sqrt{k + \sqrt{k + \cdots + \sqrt{k}}},}$$

  with exactly n square roots.

  • (a) Show that (a _n ) _n ≥ 1 is convergent.

  • (b) Show that, if k is odd, then the limit of the sequence is an irrational number.

  • (c) Find all natural values of k for which sequence converges to an integer.

• 19. For each positive real a, let the sequence (a _n ) _n ≥ 1 be defined by a ₁ = 1 and

  $$\displaystyle{a_{k+1} = \frac{1} {2}\left (a_{k} + \frac{a} {a_{k}}\right ),}$$

  for every integer k ≥ 1. Prove that the sequence converges to $\sqrt{a}$.

• 20. (TT) The set of natural numbers is partitioned into m disjoint, infinite and nonconstant arithmetic progressions, of common ratios d ₁, d ₂, …, d _m . Prove that

  $$\displaystyle{ \frac{1} {d_{1}} + \frac{1} {d_{2}} + \cdots + \frac{1} {d_{m}} = 1.}$$

• 21. (OIMU) Let c and α be positive real constants⁴ and Q be a square in the plane. Prove that there doesn’t exist a surjection f: [0, 1] → Q for which

  $$\displaystyle{d(f(x),f(y)) \leq c\vert x - y\vert ^{\alpha +1/2}}$$

  for all 0 ≤ x, y ≤ 1, where we let $d(A,B) =\, \overline{AB}$ denote the euclidean distance between the points A and B in the plane.

• 22. (Turkey) Let (a _n ) _n ≥ 1 be a sequence of integers such that $0 < a_{n+1} - a_{n} < \sqrt{a_{n}}$, for every natural n. Given real numbers x and y, with 0 ≤ x < y ≤ 1, prove that there exist natural numbers m and n such that

  $$\displaystyle{x < \frac{a_{m}} {a_{n}} < y.}$$

• 23. (IMO shortlist) Let (a _n ) _n ≥ 1 be a sequence of positive reals. Show that

  $$\displaystyle{1 + a_{n} > a_{n-1}\root{n}\of{2}}$$

  for infinitely many values of n.

7.3 Kronecker’s Lemma

In this section, we apply some of the ideas exposed so far in this chapter to study the important concept of dense set, as well as to present an interesting geometric application of it. We start by recalling the following definition.

Definition 7.29

Given an interval I, we say that a subset X of I is dense (in I ) if, for every a ∈ I and ε > 0, it happens that

$$\displaystyle{X \cap (a-\epsilon,a+\epsilon )\neq \emptyset.}$$

Intuitively, the density of X in I means that X is spreaded all over I. Problem 4, page 206, shows that both $\mathbb{Q}$ and $\mathbb{R}\setminus \mathbb{Q}$ are dense in $\mathbb{R}$, whereas Problem 5, page 206, shows that the set of dyadic rationals, i.e., rational numbers of the form $\frac{n} {2^{k}}$, where $n,k \in \mathbb{Z}_{+}$ are such that 0 ≤ n ≤ 2^k, is dense in [0, 1].

A quite useful result on the density (in $\mathbb{R}$) of certain of its subsets is the content of Theorem 7.31 and Corollary 7.32, which are collectively known in the mathematical literature as Kronecker’s lemma .⁵ The proofs we present, albeit not being the simplest ones, have the advantage of deriving from a circle of ideas which are interesting in themselves. First of all, we need yet another definition.

Definition 7.30

A nonempty subset G of $\mathbb{R}$ is said to be an additive subgroup of $\mathbb{R}$ if, for all x, y ∈ G, we have x − y ∈ G.

Evidently, {0}, $\mathbb{Z}$, $\mathbb{Q}$ and $\mathbb{R}$ itself are additive subgroups of $\mathbb{R}$. For a less obvious example, given real numbers x ₁, …, x _k , it is immediate to verify (see Problem 1) that the set

$$\displaystyle{ G_{x_{1},\ldots,x_{k}} =\{ a_{1}x_{1} + \cdots + a_{k}x_{k};\,a_{1},\ldots,a_{k} \in \mathbb{Z}\} }$$

(7.1)

is also an additive subgroup of $\mathbb{R}$.

Now, let G be an arbitrary additive subgroup of $\mathbb{R}$ and take x ∈ G. By the above definition, we have 0 = x − x ∈ G. Thus, for x, y ∈ G, we also have − y = 0 − y ∈ G and, hence, x + y = x − (−y) ∈ G; therefore, G is closed under the operation of addition.

Hence, if α ∈ G, then 2α = α +α ∈ G; moreover, if k α ∈ G, for some $k \in \mathbb{N}$, then (k + 1)α = k α +α ∈ G, so that m α ∈ G, for every $m \in \mathbb{N}$. Since 0α = 0 ∈ G and (−k)α = −k α ∈ G for every $k \in \mathbb{N}$, it follows that

$$\displaystyle{ G_{\alpha } =\{ m\alpha;\,m \in \mathbb{Z}\} \subset G. }$$

(7.2)

The coming result collects two central facts on additive subgroups of $\mathbb{R}$.

Theorem 7.31 (Kronecker)

Let G ≠ {0} be an additive subgroup of $\mathbb{R}$ , and let $G_{+}^{{\ast}} = G \cap \mathbb{R}_{+}^{{\ast}}$ .

• (a) If inf (G ₊ ^∗ ) = 0, then G is dense in $\mathbb{R}$ .

• (b) If inf (G ₊ ^∗ ) = α > 0, then α ∈ G and G = G _α .

Proof

  

• (a) Suppose inf(G ₊ ^∗) = 0, let $a \in \mathbb{R}$ and ε > 0 be given. We have to show that G ∩ (a −ε, a +ε) ≠ ∅. Since x ∈ G ⇔ −x ∈ G, it suffices to analyse the case a ≥ 0. If a −ε < 0, we have 0 ∈ G ∩ (a −ε, a +ε) and there is nothing to do. Suppose, then, that a −ε ≥ 0.

  The hypothesis inf(G ₊ ^∗) = 0 guarantees the existence of x ∈ G ₊ ^∗ such that x < 2ε. Letting m be the greatest nonnegative integer such that mx ≤ a −ε, we claim that (m + 1)x ∈ G ∩ (a −ε, a +ε). Indeed, if (m + 1)x ≥ a +ε, we would have

  $$\displaystyle{mx \leq a-\epsilon < a+\epsilon \leq (m + 1)x,}$$

  so that

  $$\displaystyle{x = (m + 1)x - mx \geq (a+\epsilon ) - (a-\epsilon ) = 2\epsilon,}$$

  thus contradicting the choice of x. Hence, (m + 1)x ∈ (a −ε, a +ε) ∩ G.

• (b) Suppose inf(G ₊ ^∗) = α > 0. We initially claim that α ∈ G ₊ ^∗. By the sake of arriving at a contradiction, suppose that α ∉ G ₊ ^∗. Then, the definition of infimum of a set would assure the existence of elements β, γ ∈ G ₊ ^∗ such that α < β < γ < 2α. However, since G is an additive subgroup of $\mathbb{R}$, it would follow from here that γ −β ∈ G ₊ ^∗, with

  $$\displaystyle{0 <\gamma -\beta < 2\alpha -\alpha =\alpha.}$$

  This contradicts the fact that α = inf(G ₊ ^∗), thus showing that α ∈ G ₊ ^∗.

  Now, take any x ∈ G ₊ ^∗ and let $q = \left \lfloor \frac{x} {\alpha } \right \rfloor$ and $r = x -\alpha \left \lfloor \frac{x} {\alpha } \right \rfloor$, so that $q \in \mathbb{Z}_{+}$, 0 ≤ r < α and x = q α + r. If r > 0, then the fact that G is an additive subgroup of $\mathbb{R}$ would imply r = x − q α ∈ G ₊ ^∗, with 0 < r < α. Since this contradicts the fact that α = inf(G ₊ ^∗), we conclude that r = 0 and, hence, x = q α ∈ G _α .

  Therefore,

  $$\displaystyle{G_{+}^{{\ast}}\subset \{ n\alpha;\,n \in \mathbb{N}\}}$$

  and, since the opposite inclusion was already established in (7.2), we actually have $G_{+}^{{\ast}} =\{ n\alpha;\,n \in \mathbb{N}\}$. Finally, since G = G ₊ ^∗∪{ 0} ∪ G ₋ ^∗, where G ₋ ^∗ = { −x; x ∈ G ₊ ^∗}, it is immediate to see that

  $$\displaystyle{G =\{ m\alpha;\,m \in \mathbb{Z}\} = G_{\alpha }.}$$

   □ 

In the coming corollary, we stick to the notation set forth in (7.1).

Corollary 7.32 (Kronecker)

If α is an irrational number, then the additive subgroup $G_{1,\alpha } =\{ m + n\alpha;\,m,n \in \mathbb{Z}\}$ of $\mathbb{R}$ is dense in $\mathbb{R}$ .

Proof

By the sake of simplicity of notation, let G = G _1, α . By the previous theorem, in order to prove that G is dense in $\mathbb{R}$, it suffices to prove that inf(G ₊ ^∗) = 0.

If this was not the case, then, once more from Kronecker’s theorem, there would exist a positive real number β such that inf(G ₊ ^∗) = β > 0 and G = G _β . Since both α and 1 +α belong to G, there would exist distinct, nonzero integers m and n for which

$$\displaystyle{\alpha = n\beta \ \ \text{and}\ \ 1+\alpha = m\beta.}$$

Now, since α is irrational, the first equality above would give n ≠ 0 and $\beta = \frac{\alpha } {n}\notin \mathbb{Q}$. On the other hand, we would also have

$$\displaystyle{(m - n)\beta = (1+\alpha )-\alpha = 1,}$$

so that $\beta = \frac{1} {m-n} \in \mathbb{Q}$.

We have then reached a contradiction, which came from the supposition that G is not dense in $\mathbb{R}$. Therefore, G is indeed dense in $\mathbb{R}$, as we wished to show. □ 

Our next corollary refines the conclusion of the previous one.

Corollary 7.33

If α is an irrational number, then the following sets are dense in  $\mathbb{R}$ :

• (a) $A =\{ m + n\alpha;\,m,n \in \mathbb{Z}\,\,\text{and}\,\,m < 0 < n\}$ .

• (b) $B =\{ m + n\alpha;\,m,n \in \mathbb{Z}\,\,\text{and}\,\,n < 0 < m\}$ .

Proof

Without any loss of generality, we can suppose that α > 0. Let’s prove item (a), the proof of item (b) being totally analogous. Given $a \in \mathbb{R}$ and ε > 0, we want to establish the existence of x ∈ A such that a −ε < x < a +ε. Suppose that a −ε ≥ 0 (the remaining cases are entirely analogous), and let δ = min{α, 2ε} > 0.

We first claim that there exist $m,n \in \mathbb{Z}$ such that m < 0 < n and m + n α ∈ A ∩ (0, δ). By contradiction, suppose that

$$\displaystyle{m + n\alpha \in A \cap (0,\delta ) \Rightarrow n \leq 0.}$$

Choose (by the former corollary) x ₀ = m ₀ + n ₀ α ∈ A ∩ (0, δ), with the greatest possible n ₀ ≤ 0. Since Kronecker’s lemma guarantees that A ∩ (0, x ₀) is infinite, we can take x ₁ = m ₁ + n ₁ α ∈ A ∩ (0, x ₀), with n ₁ < n ₀. Then,

$$\displaystyle{x_{0} - x_{1} = (m_{0} - m_{1}) + (n_{0} - n_{1})\alpha \in A \cap (0,x_{0}) \subset A \cap (0,\delta ),}$$

which is a contradiction, for, n ₀ − n ₁ > 0. Hence, we can choose m + n α ∈ A ∩ (0, δ), with n > 0. This being said, if it were m ≥ 0, we would have m + n α ≥ α ≥ δ, which is another contradiction. Therefore, m < 0 and, thus, A ∩ (0, δ) ≠ ∅.

Now, take x ∈ A ∩ (0, δ) and consider all numbers of the form kx, with $k \in \mathbb{Z}_{+}$. Letting k ₀ be the greatest nonnegative integer for which k ₀ x ≤ a −ε, we claim that (k ₀ + 1)x ∈ (a −ε, a +ε). Indeed, if it were (k ₀ + 1)x ≥ a +ε, we would have

$$\displaystyle{k_{0}x \leq a-\epsilon < a+\epsilon \leq (k_{0} + 1)x}$$

and, hence,

$$\displaystyle{\delta > x = (k_{0} + 1)x - k_{0}x \geq (a+\epsilon ) - (a-\epsilon ) = 2\epsilon.}$$

This contradicts the choice of δ. □ 

The discussion of the following example uses a few simple facts on plane Euclidean Geometry, for which we refer the reader to [4].

Example 7.34 (Brazil)

Let $\Pi$ be an euclidean plane and $f: \Pi \rightarrow \Pi$ be a function such that

$$\displaystyle{d(P,Q) = 1 \Rightarrow d(f(P),f(Q)) = 1,}$$

for all $P,Q \in \Pi$. Prove that f is an isometry of $\Pi$, i.e., prove that, for all $P,Q \in \Pi$, one has d(P, Q) = d(f(P), f(Q)), where $d(X,Y ) =\, \overline{XY }$ stands for the euclidean distance between the points X and Y.

Proof

For $P \in \Pi$, we let f(P) be systematically denoted by P ^′, so that $d(P,Q) =\, \overline{PQ}$ and $d(f(P),f(Q)) =\, \overline{P^{{\prime}}Q^{{\prime}}}$. Firstly, let’s show that f must preserve segments of length $\sqrt{3}$.

Claim 1

$$\displaystyle{\,\overline{PQ} = \sqrt{3} \Rightarrow \,\overline{P^{{\prime}}Q^{{\prime}}} = \sqrt{3}.}$$

Indeed, given points P and Q in the plane for which $\,\overline{PQ} = \sqrt{3}$, let’s construct points R and S such that both QRS and PRS are equilateral triangles whose side lengths are equal to 1 (cf. Fig. 7.1).

Fig. 7.1

$\,\overline{PQ} = \sqrt{3} \Rightarrow \,$ $\overline{P^{{\prime}}Q^{{\prime}}} = \sqrt{3}$

Let us counterclockwise rotate rhombus PRQS with center at P, until we get a rhombus PTUV such that $\,\overline{QU} = 1$.

Observe that the images P ^′, R ^′ and S ^′ of P, R and S form an equilateral triangle of side lengths equal to 1. Since $\,\overline{Q^{{\prime}}R^{{\prime}}} =\, \overline{Q^{{\prime}}S^{{\prime}}} = 1$, it follows that Q ^′ = P ^′ or P ^′ R ^′ Q ^′ S ^′ is a rhombus congruent to PRQS (so that $\,\overline{P^{{\prime}}Q^{{\prime}}} = \sqrt{3}$). In order to discard the first possibility, it suffices to note that, if it were P ^′ = Q ^′, then T ^′, U ^′ and V ^′ would all be points on a circle centered at P ^′ = Q ^′, while being vertices of an equilateral triangle of side lengths equal to 1, which is an absurd.

Claim 2

For every positive integer n, we have

$$\displaystyle{\,\overline{PQ} = n \Rightarrow \,\overline{P^{{\prime}}Q^{{\prime}}} = n.}$$

It suffices to establish the case n = 2, the general case being totally analogous. Let P and Q be such that $\,\overline{PQ} = 2$, and let R be the midpoint of PQ, such that $\,\overline{PR} =\, \overline{RQ} = 1$ (cf. Fig. 7.2).

Fig. 7.2

$\,\overline{PQ} = 2 \Rightarrow \,$ $\overline{P^{{\prime}}Q^{{\prime}}} = 2$

Let’s consider points S and T such that PRS, RST and QRT are equilateral triangles of side lengths equal to 1, all situated on one of the half-planes determined by line $\,\stackrel{\longleftrightarrow }{PQ}$. Using claim 1 twice, it’s immediate that $\,\overline{P^{{\prime}}Q^{{\prime}}} = 2$.

Analogously, we can prove that

$$\displaystyle{\,\overline{PQ} = n\sqrt{3} \Rightarrow \,\overline{P^{{\prime}}Q^{{\prime}}} = n\sqrt{3}.}$$

Claim 3

$\,\overline{P^{{\prime}}Q^{{\prime}}}\geq \,\overline{PQ}$, for all points P and Q in the plane.

In order to prove this claim, let $\,\overline{PQ} = l$, such that l is neither a natural number nor a real number of the form $n\sqrt{3}$, for some $n \in \mathbb{N}$. By Corollary 7.33 (see, also, Problem 3), we can take sequences (m _k ) _k ≥ 1 e (n _k ) _k ≥ 1 of integers satisfying the following conditions:

• i. m _k  < 0 < n _k , for every k ≥ 1;

• ii. $\lim _{k\rightarrow +\infty }(m_{k} + n_{k}\sqrt{3}) = l$;

• iii. $\max \{0,l - 1\} < m_{k} + n_{k}\sqrt{3} < l$, for every k ≥ 1.

Let’s first show that there exists a triangle of side lengths l, − m _k and $n_{k}\sqrt{3}$. To this end, since $m_{k} + n_{k}\sqrt{3} < l$, we have $l + (-m_{k}) > n_{k}\sqrt{3}$; also, $l + (m_{k} + n_{k}\sqrt{3}) > 0$, which implies $l + n_{k}\sqrt{3} > -m_{k}$; finally, from $l - 1 < m_{k} + n_{k}\sqrt{3}$ we get

$$\displaystyle{n_{k}\sqrt{3} + (-m_{k}) > n_{k}\sqrt{3} + m_{k} + 1 > l.}$$

Therefore, triangle inequality assures the existence of a point $R \in \Pi \setminus \,\stackrel{\longleftrightarrow }{PQ}$ such that $\,\overline{PR} = n_{k}\sqrt{3}$ and $\,\overline{RQ} = -m_{k}$; since we already have $\,\overline{PQ} = l$, there is nothing left to do.

It follows from what we did above (cf. Fig. 7.3) that $\,\overline{P^{{\prime}}Q^{{\prime}}} +\, \overline{R^{{\prime}}Q^{{\prime}}}\geq \,\overline{P^{{\prime}}R^{{\prime}}}$ or, which is the same, $\,\overline{P^{{\prime}}Q^{{\prime}}}\geq \,\overline{P^{{\prime}}R^{{\prime}}}-\,\overline{R^{{\prime}}Q^{{\prime}}}$. However, since $\,\overline{P^{{\prime}}R^{{\prime}}} = n_{k}\sqrt{3}$ and $\,\overline{R^{{\prime}}Q^{{\prime}}} = -m_{k}$, we get $\,\overline{P^{{\prime}}Q^{{\prime}}} \geq n_{k}\sqrt{3} + m_{k}$. On the other hand, since $n_{k}\sqrt{3} + m_{k}\stackrel{k}{\longrightarrow }l$, it comes that $\,\overline{P^{{\prime}}Q^{{\prime}}}\geq l =\, \overline{PQ}$.

Fig. 7.3

$\,\overline{P^{{\prime}}Q^{{\prime}}}\geq \,\overline{PQ}$

Let’s consider again points P and Q in the plane, with $\,\overline{PQ} = l$. Tesselate the plane with equilateral triangles of side lengths all equal to 1, such that one of these triangles has one of its vertices at P and one of its sides on line $\,\stackrel{\longleftrightarrow }{PQ}$ (cf. Fig. 7.4).

Fig. 7.4

$\,\overline{PQ} = l \Rightarrow \,$ $\overline{P^{{\prime}}Q^{{\prime}}} = l$

By what we did above, the images by f of the vertices of such a triangulation form the vertices of an analogous triangulation. On the other hand, if X is an arbitrary vertex of the original triangulation (see Fig. 7.4 once more), then $\,\overline{X^{{\prime}}Q^{{\prime}}} \geq \,\overline{XQ}$, for every point Q of the plane. Geometrically, this means that Q ^′ doesn’t belong to the interior of the disk centered at X ^′ and having radius $\,\overline{XQ}$. However, since this is true for every vertex of the triangulation, we must necessarily have $\,\overline{P^{{\prime}}Q^{{\prime}}} = l$. □ 

Apart from the following problems, other interesting applications of Kronecker’s lemma will appear in the context of continuity in Problem 14, page 264, as well as in Sect. 10.8 (cf. Examples 10.59 and 10.60), when we have at our disposal the concepts and elementary properties of logarithms.

Problems: Section 7.3

• 1. * Given real numbers x ₁, …, x _k , verify that the set

  $$\displaystyle{G_{x_{1},\ldots,x_{k}} =\{ a_{1}x_{1} + \cdots + a_{k}x_{k};\,a_{1},\ldots,a_{k} \in \mathbb{Z}\}}$$

  is indeed an additive subgroup of $\mathbb{R}$.

  For the next problem, the reader might want to recall the definitions of integer part and fractional part of a real number, given in Problems 9 and 10, page 152.

• 2. The purpose of this problem is to give a direct proof of Corollary 7.32. To this end, given $\alpha \in \mathbb{R}\setminus \mathbb{Q}$, $a \in \mathbb{R}$ and ε > 0, start by choosing $p \in \mathbb{N}$ such that $\frac{1} {p} < 2\epsilon$ and do the following items:

  • (a) Show that at least two of the numbers {α}, {2α}, …, {(p + 1)α} belong to a single interval of the form $\big[\frac{k} {p}, \frac{k+1} {p} \big)$, for some integer k satisfying 0 ≤ k < p.

  • (b) Use the result of (a) to show that there exist $m^{{\prime}},n^{{\prime}}\in \mathbb{Z}$ such that $0 < m^{{\prime}} + n^{{\prime}}\alpha < \frac{1} {p}$.

  • (c) Use the result of (b) to show that there exists $r \in \mathbb{Z}$ such that r(m ^′ + n ^′ α) ∈ (a −ε, a +ε).

• 3. * Given $\alpha,l \in \mathbb{R}$, with α irrational, show that there exist sequences (m _k ) _k ≥ 1 and (n _k ) _k ≥ 1 of integers satisfying the following conditions:

  • (a) m _k  < 0 < n _k , for every k ≥ 1;

  • (b) lim _{k → +∞} (m _k + n _k α) = l.

  For the next problem, we assume from the reader some familiarity with the basics of plane analytic geometry and vector algebra in the plane. We refer to Chaps. 6 and 8 of [4] for the necessary background.

• 4. A subset X of an euclidean plane $\Pi$ is said to be dense in π provided X intersects every disk of $\Pi$. Now, let O be a fixed point in $\Pi$. A nonempty subset X of $\Pi$ is said to be an additive subgroup of $\Pi$ with respect to O provided the following condition is satisfied: for every A, B ∈ X, if $\overrightarrow{OA} -\overrightarrow{ OB} =\overrightarrow{ OC}$, then C ∈ X.

  • (a) If $A_{1},\ldots,A_{n} \in \Pi$ and

    $$\displaystyle{X_{A_{1},\ldots,A_{n}} =\Big\{\sum _{ k=1}^{n}m_{ k}\overrightarrow{OA_{k}};\,m_{k} \in \mathbb{Z},\,\,\forall \,\,1 \leq k \leq n\Big\},}$$

    show that $X_{A_{1},\ldots,A_{n}}$ is an additive subgroup of $\Pi$ with respect to O, which is not dense in it.

  • (b) Choose A and B in $\Pi$ such that $O\notin \,\stackrel{\longleftrightarrow }{AB}$. If $\alpha \in \mathbb{R}\setminus \mathbb{Q}$ and

    $$\displaystyle{Y _{A,B} =\{ m\overrightarrow{OA} + (n + p\alpha )\overrightarrow{OB};\,\,m,n,p \in \mathbb{Z}\},}$$

    show that Y _A, B is an additive subgroup of $\Pi$ with respect to O, which is not dense in it.

  • (c) Give an example of an additive subgroup X of $\Pi$ with respect to O, different from $\Pi$ itself and dense in it.

  • (d) Let X be an additive subgroup of $\Pi$ with respect to O. Do the following items:

    • i. Given a line r through O, look at it as a real line, with O representing 0. Prove that X ∩ r is either the empty set or an additive subgroup of r.

    • ii. If there exist distinct lines r and s through O such that X ∩ r, X ∩ s ≠ ∅, prove that X is dense in $\Pi$.

    • iii. If X is not dense in $\Pi$ and is not contained in a single line, prove that there exists a direction d in $\Pi$ such that X is contained in the union of a family of equally spaced lines, all parallel to d.

  Although the next problem does not really use Kronecker’s lemma, this is the best place to put it.

• 5. Let α and β be positive irrationals such that $\frac{1} {\alpha } + \frac{1} {\beta } = 1$. Our purpose is to show that the sets

  $$\displaystyle{\left \{\lfloor k\alpha \rfloor;\,\,k \in \mathbb{N}\right \}\ \ \text{and}\ \ \left \{\lfloor k\beta \rfloor;\,\,k \in \mathbb{N}\right \}}$$

  form a partition of the natural numbers. To this end, do the following items:

  • (a) Show that if α > 1 is irrational and $n \in \mathbb{N}$, then n is a term of the sequence (⌊k α⌋) _k ≥ 1 if and only if $\Big\{\frac{n} {\alpha } \Big\} > 1 -\frac{1} {\alpha }$.

  • (b) Given $n \in \mathbb{N}$, show that $\Big\{\frac{n} {\alpha } \Big\} +\Big\{ \frac{n} {\beta } \Big\} = 1$.

  • (c) Conclude that either $\Big\{\frac{n} {\alpha } \Big\} > 1 -\frac{1} {\alpha }$ or $\Big\{\frac{n} {\beta } \Big\} > 1 -\frac{1} {\beta }$, but not both.

  • (d) Finish the proof.

  In the notations above, one says that (a _k ) _k ≥ 1 and (b _k ) _k ≥ 1, given by a _k  = ⌊k α⌋ and b _k  = ⌊k β⌋, are the Beatty sequences ⁶ corresponding to α and β.

7.4 Series of Real Numbers

Let (a _n ) _n ≥ 1 be a sequence of real numbers. By the series

$$\displaystyle{\sum _{n=1}^{+\infty }a_{ n},}$$

or simply ∑ _n ≥ 1 a _n , we mean the sequence (s _n ) _n ≥ 1, where s _n  = a ₁ + a ₂ + ⋯ + a _n for n ≥ 1. The real number s _n is called the n-th partial sum of the series ∑ _n ≥ 1 a _n , and we say that such a series converges to $s \in \mathbb{R}$ if the sequence (s _n ) _n ≥ 1 of its partial sums converges to s. In this case, we say that s is the sum of the series and write

$$\displaystyle{ \sum _{n\geq 1}a_{n} = s. }$$

(7.3)

In other words, whenever we write ∑ _k ≥ 1 a _k  = s, we will be saying that the terms of the sequence of finite sums s _n  = a ₁ + a ₂ + ⋯ + a _n come closer and closer to the real number s, as long as n → +∞. It is in this sense that equality (7.3) must be thought of, as a limit.

We shall sometimes have a sequence (a _n ) _n ≥ 0 of reals, in which case the corresponding series will be denoted by ∑ _n ≥ 0 a _n . We leave to the reader the (immediate) task of adapting the former and coming discussions to such a situation.

Our main interest in this section is to find out efficient criteria to decide whether a given series converges or not. If it doesn’t converge, we shall say that it is a divergent series. Let’s see two simple examples of divergent series.

Example 7.35

The series ∑ _k ≥ 1 k and ∑ _k ≥ 1(−1)^k diverge.

Proof

The first series diverges, for its n-th partial sum is $s_{n} = 1 + 2 + \cdots + n = \frac{n(n+1)} {2}$, so that (s _n ) _n ≥ 1 is a divergent sequence. In the second case, the n-th partial sum s _n of the given series is such that s _n  = 0 if n is even and s _n  = −1 if n is odd, so that (s _n ) _n ≥ 1 is also a divergent sequence. □ 

Given a series ∑ _n ≥ 1 a _n , we refer to a generic term a _n as the general term of the series. The following proposition gives a necessary condition on the general term of a series, if it is to converge.

Proposition 7.36

If the series ∑ _k≥1 a _k converges, then $a_{k}\stackrel{n}{\longrightarrow }0$ .

Proof

Given ε > 0, we want to prove that there exists $n_{0} \in \mathbb{N}$ such that n > n ₀ ⇒ | a _n  |  < ε. To this end, let l = ∑ _k ≥ 1 a _k . By the definition of convergence for series, there exists $n_{0} \in \mathbb{N}$ such that

$$\displaystyle{n \geq n_{0} \Rightarrow \vert (a_{1} + a_{2} + \cdots + a_{n}) - l\vert < \frac{\epsilon } {2}.}$$

Therefore, it follows from the triangle inequality that, for n > n ₀, we have

$$\displaystyle{ \begin{array}{rl} \vert a_{n}\vert &\, \leq \vert (a_{1} + a_{2} + \cdots + a_{n}) - l\vert + \vert l - (a_{1} + a_{2} + \cdots + a_{n-1})\vert \\ &\, \leq \frac{\epsilon } {2} + \frac{\epsilon } {2} =\epsilon.\end{array} }$$

 □ 

The converse to the above proposition is not true, namely, there are divergent series ∑ _k ≥ 1 a _k for which $a_{k}\stackrel{n}{\longrightarrow }0$. The classical example is that of the harmonic series , i.e., the series $\sum _{k\geq 1} \frac{1} {k}$, whose divergence is established in the coming example and will find further use in these notes.

Example 7.37

Given $n \in \mathbb{N}$, let m be the only natural number such that 2^m ≤ n < 2^m+1. Then,

$$\displaystyle{ \sum _{k=1}^{n}\frac{1} {k} \geq \frac{m} {2} + 1. }$$

(7.4)

In particular, the harmonic series diverges.

Proof

Note that, for every integer k > 1,

$$\displaystyle{ \frac{1} {2^{k-1} + 1} + \frac{1} {2^{k-1} + 2} + \cdots + \frac{1} {2^{k}} >\mathop{\underbrace{ \frac{1} {2^{k}} + \frac{1} {2^{k}} + \cdots + \frac{1} {2^{k}}}}\limits _{2^{k-1}\,\mathrm{times}} = \frac{1} {2}.}$$

Hence,

$$\displaystyle{ \begin{array}{rl} 1 + \frac{1} {2} +\sum _{ j=3}^{n}\frac{1} {j} &\, \geq 1 + \frac{1} {2} +\sum _{ j=3}^{2^{m} }\frac{1} {j} \\ &\, = 1 + \frac{1} {2} +\sum _{ k=2}^{m}\left ( \frac{1} {2^{k-1}+1} + \cdots + \frac{1} {2^{k}}\right ) \\ &\, > 1 + \frac{1} {2} +\sum _{ k=2}^{m}\frac{1} {2} = 1 + \frac{m} {2}. \end{array} }$$

 □ 

In what comes next, we shall show that, for every rational r > 1, the series $\sum _{k\geq 1} \frac{1} {k^{r}}$ converges. To this end, we need to examine the convergence of a geometric series , i.e., of a series of the form

$$\displaystyle{\sum _{k\geq 1}q^{k-1},}$$

for a certain nonzero real number q. In this sense, we have the following important result.

Proposition 7.38

Given $q \in \mathbb{R}\setminus \{0\}$ , the geometric series ∑ _k≥1 q ^k−1 converges if and only if 0 < |q| < 1. Moreover, if this is so, then

$$\displaystyle{\sum _{k\geq 1}q^{k-1} = \frac{1} {1 - q}.}$$

Proof

If | q | ≥ 1, the geometric series diverges, since its general term q ^k−1 doesn’t converge to 0. Suppose, then, that 0 <  | q |  < 1, and let s _n  = 1 + q + ⋯ + q ⁿ⁻¹ be the n-th partial sum of the series. By the formula for the sum of the terms of a finite GP, we have

$$\displaystyle{s_{n} = \frac{1 - q^{n}} {1 - q} = \frac{1} {1 - q} - \frac{q^{n}} {1 - q}.}$$

Hence, in order to show that the series converges to $\frac{1} {1-q}$, it suffices to show that $q^{n}\stackrel{n}{\longrightarrow }0$. But this was done in Example 7.12. □ 

We can now discuss the promised example.

Example 7.39

If r > 1 is rational, then the series $\sum _{k\geq 1} \frac{1} {k^{r}}$ converges.⁷

Proof

By the Bolzano-Weierstrass theorem, it suffices to show that the sequence (s _n ) _n ≥ 1 of the partial sums $s_{n} =\sum _{ k=1}^{n} \frac{1} {k^{r}}$ is bounded. To this end, given $n \in \mathbb{N}$, take $m \in \mathbb{N}$ such that 2^m > n. Then,

$$\displaystyle{ \begin{array}{rl} s_{n}&\, \leq 1 + \left ( \frac{1} {2^{r}} + \frac{1} {3^{r}}\right ) + \cdots + \left ( \frac{1} {(2^{m-1})^{r}} + \cdots + \frac{1} {(2^{m}-1)^{r}}\right ) \\ &\, < 1 + 2 \cdot \frac{1} {2^{r}} + 4 \cdot \frac{1} {4^{r}} + \cdots + 2^{m-1} \cdot \frac{1} {2^{(m-1)r}} \\ & \, < 1 + \frac{1} {2^{r-1}} + \frac{1} {4^{r-1}} + \cdots + \frac{1} {2^{(m-1)(r-1)}} \\ & \, <\sum _{k\geq 0} \frac{1} {2^{(r-1)k}}. \end{array} }$$

However, since r > 1, we have $0 < \frac{1} {2^{r-1}} < 1$, and it follows from the previous proposition that

$$\displaystyle{\sum _{k\geq 0} \frac{1} {2^{(r-1)k}} = \frac{2^{r-1}} {2^{r-1} - 1}.}$$

Therefore, we conclude that $0 < s_{n} < \frac{2^{r-1}} {2^{r-1}-1}$ for every $n \in \mathbb{N}$, so that the sequence (s _n ) _n ≥ 1 is, indeed, bounded. □ 

Remark 7.40

For the sake of curiosity, we inform the reader that $\sum _{k\geq 1} \frac{1} {k^{r}} =\zeta (r)$, where $\zeta: (1,+\infty ) \rightarrow \mathbb{R}$ stands for the famous Riemann’s zeta function .⁸ An elementary computation of $\zeta (2) = \frac{\pi ^{2}} {6}$ will be hinted to at Problem 12, page 470. We also refer to Chap. 9 of [5]. Note, however, that for an odd natural number m > 1 the computation of the exact numerical value of ζ(m) is an open problem in Mathematics.

We finish our initial list of examples of series with an additional application of Bolzano-Weierstrass theorem to the convergence of series. In the coming example, we introduce one of the most important constants of Mathematics, the number e, which will play a major role in Sect. 10.7

Example 7.41

The series $\sum _{k\geq 0} \frac{1} {k!}$ converges to an irrational number e, such that 2 < e < 3. In symbols,

$$\displaystyle{ e =\sum _{k\geq 0} \frac{1} {k!}. }$$

(7.5)

Proof

Let (s _n ) _n ≥ 0 be the sequence of the partial sums of the given series, i.e.,

$$\displaystyle{s_{n} = 1 + \frac{1} {1!} + \frac{1} {2!} + \cdots + \frac{1} {n!}.}$$

For this sequence, we clearly have 1 = s ₀ < s ₁ < s ₂ < ⋯ ; on the other hand, since k!  > 2^k−1 for every integer k > 2, we have, for an integer n ≥ 4,

$$\displaystyle{s_{n} =\sum _{ k=0}^{n} \frac{1} {k!} < 1 + 1 + \frac{1} {2} + \frac{1} {6} +\sum _{ k=4}^{n} \frac{1} {2^{k-1}} < \frac{8} {3} +\sum _{k\geq 4} \frac{1} {2^{k-1}} = \frac{35} {12},}$$

where we used the formula for the sum of a geometric series in the last equality above. Hence, the sequence (s _n ) _n ≥ 0 is increasing and bounded, thus, convergent. Now, item (b) of Proposition 7.15, together with s ₂ = 2 and $s_{n} < \frac{35} {12}$ for every integer n ≥ 4, gives 2 < e < 3 (for an approximation of e with five correct decimal places, see Problem 8).

We shall now show that e is irrational.⁹ To this end, observe initially that, for natural numbers 1 < n < m, we have

$$\displaystyle\begin{array}{rcl} s_{m} - s_{n}& \,=& \sum _{k=n+1}^{m} \frac{1} {k!} <\sum _{k\geq n+1} \frac{1} {k!} \\ & \,=& \frac{1} {(n + 1)!}\left (1 + \frac{1} {n + 2} + \frac{1} {(n + 2)(n + 3)} + \cdots \,\right ) \\ & \,<& \frac{1} {(n + 1)!}\sum _{k\geq 0} \frac{1} {(n + 2)^{k}} = \frac{1} {(n + 1)!} \cdot \frac{n + 2} {n + 1},{}\end{array}$$

(7.6)

where, once more, we used the formula for the sum of a geometric series in the last equality above.

Therefore, yet for natural numbers 1 < n < m, it follows from the above computations that

$$\displaystyle{s_{m} = s_{n} +\sum _{ k=n+1}^{m} \frac{1} {k!} < s_{n} + \frac{1} {(n + 1)!} \cdot \frac{n + 2} {n + 1},}$$

and item (a) of Proposition 7.15 gives

$$\displaystyle{e =\lim _{m\rightarrow +\infty }s_{m} \leq s_{n} + \frac{1} {(n + 1)!} \cdot \frac{n + 2} {n + 1}.}$$

Thus,

$$\displaystyle{s_{n} < e \leq s_{n} + \frac{1} {(n + 1)!} \cdot \frac{n + 2} {n + 1}.}$$

Multiplying this last inequality by (n − 1)! and noticing that $s_{n} = s_{n-1} + \frac{1} {n!}$, we conclude that

$$\displaystyle{(n - 1)!s_{n-1} + \frac{1} {n} < (n - 1)!e \leq (n - 1)!s_{n-1} + \frac{1} {n} + \frac{n + 2} {n(n + 1)^{2}},}$$

Writing $t_{n} = n!s_{n} \in \mathbb{N}$ and observing that

$$\displaystyle{ \begin{array}{rl} \frac{1} {n} + \frac{n+2} {n(n+1)^{2}} & \, = \frac{1} {n} + \frac{1} {(n+1)^{2}} + \frac{2} {n(n+1)^{2}} \\ & \, \leq \frac{1} {3} + \frac{1} {4^{2}} + \frac{2} {3\cdot 4^{2}} = \frac{21} {48} < 1, \end{array} }$$

for every integer n > 2, we finally arrive at the estimate

$$\displaystyle{t_{n-1} < (n - 1)!e < t_{n-1} + 1,}$$

which is valid for every integer n > 2.

Now, suppose that $e = \frac{p} {q}$, with $p,q \in \mathbb{N}$. Making n = q + 1 > 2 in the above inequalities, we would get

$$\displaystyle{t_{q} < (q - 1)!p < t_{q} + 1,}$$

with $t_{q} \in \mathbb{N}$. This is a contradiction. □ 

As an alternative to (7.5), we have the following result, which will be quite useful in Sect. 10.7

Theorem 7.42

$e =\lim _{n\rightarrow +\infty }\left (1 + \frac{1} {n}\right )^{n}$ .

Proof

Let $a_{n} = \left (1 + \frac{1} {n}\right )^{n}$. Arguing as in the proof of Example 7.23, we get

$$\displaystyle{ \begin{array}{rl} a_{n}&\, = 2 +\sum _{ k=2}^{n}{n\choose k} \frac{1} {n^{k}} \\ & \, = 2 +\sum _{ k=2}^{n} \frac{1} {k!} \cdot \frac{n(n-1)(n-2)\ldots (n-k+1)} {n^{k}} \\ & \, = 2 +\sum _{ k=2}^{n} \frac{1} {k!}\left (1 - \frac{1} {n}\right )\left (1 - \frac{2} {n}\right )\ldots \left (1 -\frac{k-1} {n} \right ) \\ &\, < 2 +\sum _{ k=2}^{n} \frac{1} {k!} =\sum _{ k=0}^{n} \frac{1} {k!} <\sum _{k\geq 0} \frac{1} {k!} = e.\end{array} }$$

The above computations also give

$$\displaystyle{ \begin{array}{rl} a_{n}&\, = 2 +\sum _{ k=2}^{n} \frac{1} {k!}\left (1 - \frac{1} {n}\right )\left (1 - \frac{2} {n}\right )\ldots \left (1 -\frac{k-1} {n} \right ) \\ &\, < 2 +\sum _{ k=2}^{n+1} \frac{1} {k!}\left (1 - \frac{1} {n+1}\right )\left (1 - \frac{2} {n+1}\right )\ldots \left (1 -\frac{k-1} {n+1}\right ) = a_{n+1}. \end{array} }$$

Therefore, (a _n ) _n ≥ 1 is monotone increasing and bounded above, and hence there exists l = lim _{n → +∞} a _n . In particular, since a _n  < e for every $n \in \mathbb{N}$, item (b) of Proposition 7.15 gives l ≤ e.

Also from the computations above, given natural numbers n > m ≥ 2 we can write

$$\displaystyle{ \begin{array}{rl} a_{n}&\, = 2 +\sum _{ k=2}^{n} \frac{1} {k!}\left (1 - \frac{1} {n}\right )\left (1 - \frac{2} {n}\right )\ldots \left (1 -\frac{k-1} {n} \right ) \\ &\, > 2 +\sum _{ k=2}^{m} \frac{1} {k!}\left (1 - \frac{1} {n}\right )\left (1 - \frac{2} {n}\right )\ldots \left (1 -\frac{k-1} {n} \right ) \\ &\, > 2 +\sum _{ k=2}^{m} \frac{1} {k!}\left (1 - \frac{1} {n}\right )\left (1 - \frac{2} {n}\right )\ldots \left (1 -\frac{m-1} {n} \right ) \\ &\, > \left (1 - \frac{1} {n}\right )\left (1 - \frac{2} {n}\right )\ldots \left (1 -\frac{m-1} {n} \right )\left (2 +\sum _{ k=2}^{m} \frac{1} {k!}\right ). \end{array} }$$

Therefore, it follows from Problem 1, page 218, we have

$$\displaystyle{ \begin{array}{rl} l =\lim _{n\rightarrow +\infty }a_{n}&\, \geq \lim _{n\rightarrow +\infty }\left (1 - \frac{1} {n}\right )\left (1 - \frac{2} {n}\right )\ldots \left (1 -\frac{m-1} {n} \right )\left (2 +\sum _{ k=2}^{m} \frac{1} {k!}\right ) \\ & = 2 +\sum _{ k=2}^{m} \frac{1} {k!} =\sum _{ k=0}^{m} \frac{1} {k!}.\end{array} }$$

However, since $m \in \mathbb{N}$ was arbitrarily chosen, we conclude that $l \geq \sum _{k=0}^{m} \frac{1} {k!}$ for every $m \in \mathbb{N}$. Then, letting m → +∞ and invoking again item (b) of Proposition 7.15, we finally obtain

$$\displaystyle{l \geq \lim _{m\rightarrow +\infty }\sum _{k=0}^{m} \frac{1} {k!} =\sum _{k\geq 0} \frac{1} {k!} = e.}$$

 □ 

Back to the general development of the theory, the next result is the analogue, for series, of Proposition 7.18, and teaches us how to operate with convergent series.

Proposition 7.43

If ∑ _k≥1 a _k and ∑ _k≥1 b _k are convergent series and c is a real number, then:

• (a) ∑ _k≥1 ca _k converges and ∑ _k≥1 ca _k = c∑ _k≥1 a _k .

• (b) ∑ _k≥1 (a _k + b _k ) converges and ∑ _k≥1 (a _k + b _k ) = ∑ _k≥1 a _k + ∑ _k≥1 b _k .

Proof

  

• (a) If s _n denotes the n-th partial sum of the series ∑ _k ≥ 1 a _k , then the n-th partial sum of the series ∑ _k ≥ 1 ca _k equals cs _n . Hence, according to item (a) of Proposition 7.18, ∑ _k ≥ 1 ca _k converges, and

  $$\displaystyle{\sum _{k\geq 1}ca_{k} =\lim _{n\rightarrow +\infty }cs_{n} = c\lim _{n\rightarrow +\infty }s_{n} = c\sum _{k\geq 1}a_{k}.}$$

• (b) If s _n and t _n are the n-th partial sums of the series ∑ _k ≥ 1 a _k and ∑ _k ≥ 1 b _k , respectively, then the n-th partial sum of the series ∑ _k ≥ 1(a _k + b _k ) equals s _n + t _n . Therefore, item (b) of Proposition 7.18 assures the convergence of this last series, with

  $$\displaystyle{\sum _{k\geq 1}(a_{k} + b_{k}) =\lim _{n\rightarrow +\infty }(s_{n} + t_{n}) =\lim _{n\rightarrow +\infty }s_{n} +\lim _{n\rightarrow +\infty }t_{n} =\sum _{k\geq 1}a_{k} +\sum _{k\geq 1}b_{k}.}$$

   □ 

A quick analysis of the arguments presented in Examples 7.39 and 7.41 gives the following more general result, known as the comparison test for the convergence of series.

Proposition 7.44

Let (a _k ) _k≥1 and (b _k ) _k≥1 be sequences of positive real numbers, such that a _k ≤ b _k for every k ≥ 1. If the series ∑ _k≥1 b _k converges, then so does the series ∑ _k≥1 a _k . Moreover,

$$\displaystyle{\sum _{k\geq 1}a_{k} \leq \sum _{k\geq 1}b_{k}.}$$

Proof

Letting s _n  = ∑ _k = 1 ⁿ a _k and t _n  = ∑ _k = 1 ⁿ b _k , it follows from 0 < a _k  ≤ b _k that 0 < s _n  ≤ t _n , for every $n \in \mathbb{N}$. Since the sequence (t _n ) _n ≥ 1 converges, it is bounded. Hence, the sequence (s _n ) _n ≥ 1 is monotonic and bounded, thus convergent, by Bolzano-Weierstrass theorem. To what is left to prove, it suffices to make n → +∞ in the inequality s _n  ≤ t _n and apply the result of Problem 1, page 218. □ 

Example 7.45

Is there a sequence (a _k ) _k ≥ 1 of positive real numbers such that both series ∑ _k ≥ 1 a _k and $\sum _{k\geq 1} \frac{1} {k^{2}a_{k}}$ converge?

Solution

Suppose there is such a series. Then, item (b) of Proposition 7.43, together with the inequality between the arithmetic and geometric means, would give us

$$\displaystyle{\sum _{k\geq 1}a_{k} +\sum _{k\geq 1} \frac{1} {k^{2}a_{k}} =\sum _{k\geq 1}\left (a_{k} + \frac{1} {k^{2}a_{k}}\right ) \geq \sum _{k\geq 1}2\sqrt{a_{k } \cdot \frac{1} {k^{2}a_{k}}} =\sum _{k\geq 1} \frac{2} {k}.}$$

Therefore, by the comparison test for series, the harmonic series would be convergent, which is an absurd. □ 

The coming example uses the theory of series to give a proof of the uncountability of $\mathbb{R}$.

Example 7.46

Problem 23, 172, shows that the family $\mathcal{F}$ of infinite subsets of $\mathbb{N}$ is uncountable. On the other hand, if A = { m ₁ < m ₂ < m ₃ < ⋯ } is such a set, then the comparison test, together with the convergence of the geometric series $\sum _{j\geq 1} \frac{1} {2^{j}}$, guarantees the convergence of the series $\sum _{k\geq 1} \frac{1} {2^{m_{k}}}$.

Let B = { n ₁ < n ₂ < n ₃ < ⋯ } be another infinite subset of $\mathbb{N}$. If we show that

$$\displaystyle{\sum _{k\geq 1} \frac{1} {2^{m_{k}}}\neq \sum _{k\geq 1} \frac{1} {2^{n_{k}}},}$$

then the correspondence $A\mapsto \sum _{k\geq 1} \frac{1} {2^{m_{k}}}$ defines an injection from $\mathcal{F}$ into $\mathbb{R}$, and this guarantees that $\mathbb{R}$ is uncountable. (Otherwise, by composing such a function with a bijection from $\mathbb{R}$ to $\mathbb{N}$, we would get an injection from $\mathcal{F}$ to $\mathbb{N}$, thus contradicting the uncountability of $\mathcal{F}$.)

What is left to do is quite similar to the proof of Example 4.12. Indeed, suppose we had

$$\displaystyle{\sum _{k\geq 1} \frac{1} {2^{m_{k}}} =\sum _{k\geq 1} \frac{1} {2^{n_{k}}}.}$$

Then,

$$\displaystyle{ \frac{1} {2^{m_{1}}} <\sum _{k\geq 1} \frac{1} {2^{m_{k}}} =\sum _{k\geq 1} \frac{1} {2^{n_{k}}} \leq \sum _{j\geq n_{1}} \frac{1} {2^{j}} = \frac{1} {2^{n_{1}-1}},}$$

so that m ₁ ≥ n ₁. By reversing the roles of the two series, we analogously conclude that m ₁ ≤ n ₁, so that m ₁ = n ₁. Thus,

$$\displaystyle{\sum _{k\geq 2} \frac{1} {2^{m_{k}}} =\sum _{k\geq 2} \frac{1} {2^{n_{k}}},}$$

and a similar reasoning gives m ₂ = n ₂. Finally, by continuing this way, we get m _k  = n _k for every k ≥ 1, so that A = B.

Back to the development of the theory, for a series ∑ _k ≥ 1 a _k with infinitely many positive and negative terms the results obtained so far say nothing about its convergence. We remedy this situation from now on, starting from the following

Definition 7.47

A series ∑ _k ≥ 1 a _k is said to be absolutely convergent provided the series ∑ _k ≥ 1 | a _k  | converges.

The usefulness of the concept of absolutely convergent series stems from the coming proposition, as well as the subsequent example.

Proposition 7.48

Every absolutely convergent series is convergent.

Proof

Let ∑ _k ≥ 1 a _k be an absolutely convergent series and, for each integer n ≥ 1, let s _n  = a ₁ + a ₂ + ⋯ + a _n and t _n  =  | a ₁ | + | a ₂ | + ⋯ + | a _n  | be the n-th partial sums of the series ∑ _k ≥ 1 a _k and ∑ _k ≥ 1 | a _k  | . Given integers m > n ≥ 1, we have

$$\displaystyle{ \begin{array}{rl} \vert s_{m} - s_{n}\vert &\, = \vert a_{n+1} + a_{n+2} + \cdots + a_{m}\vert \\ &\,\leq \vert a_{n+1}\vert + \vert a_{n+2}\vert + \cdots + \vert a_{m}\vert \\ &\, = t_{m} - t_{n}. \end{array} }$$

Since (t _n ) _n ≥ 1 converges, it is a Cauchy sequence; hence, given ε > 0, there exists $n_{0} \in \mathbb{N}$ such that m > n > n ₀ ⇒ | t _m − t _n  |  < ε. With these ε and n ₀, it follows from the above inequality that

$$\displaystyle{m > n > n_{0} \Rightarrow \vert s_{m} - s_{n}\vert \leq t_{m} - t_{n} <\epsilon,}$$

and (s _n ) _n ≥ 1 is also a Cauchy sequence. Therefore, Theorem 7.27 guarantees the convergence of the sequence (s _n ) _n ≥ 1, as we wished to show. □ 

The converse to the previous proposition is not valid, namely, there are convergent series which are not absolutely convergent. The classical example is a direct application of the coming result, which is known in the mathematical literature as the Leibniz criterion ¹⁰ for the convergence of alternate series .

Proposition 7.49 (Leibniz)

If (a _n ) _n≥1 is a nonincreasing sequence of positive reals such that a _n → 0, then the alternate series ∑ _k≥1 (−1) ^k−1 a _k converges.

Proof

For each $n \in \mathbb{N}$, let s _n  = a ₁ + a ₂ + ⋯ + a _n . Condition a ₁ ≥ a ₂ ≥ a ₃ ≥ ⋯ > 0 easily gives

$$\displaystyle{ s_{1} \geq s_{3} \geq s_{5} \geq \cdots \geq s_{6} \geq s_{4} \geq s_{2}. }$$

(7.7)

On the other hand, for each $m \in \mathbb{N}$ we have

$$\displaystyle{\vert s_{2m-1} - s_{2m}\vert = a_{2m} \rightarrow 0,}$$

which clearly guarantees, in conjunction with (7.7), that (s _n ) _n ≥ 1 is a Cauchy sequence. Therefore, (s _n ) _n ≥ 1 is convergent, as desired. □ 

Example 7.50

The alternate series $\sum _{k\geq 1}\frac{(-1)^{k-1}} {k}$ converges, by a simple application of the former proposition (we shall compute its value in Problem 8, page 484). Nevertheless, the series formed by the absolute values of its terms (the harmonic series) diverges.

We now discuss quite a useful criterion for the convergence of series of nonzero real numbers. It is based on the asymptotic behavior ¹¹ of the quotient of neighboring terms of the series, and is known as the ratio test .

Proposition 7.51

Let (a _n ) _n≥1 be a sequence of nonzero real numbers, such that $\left \vert \frac{a_{n+1}} {a_{n}} \right \vert \rightarrow l$ . If l < 1, then the series ∑ _k≥1 a _k is absolutely convergent; if l > 1, then the series ∑ _k≥1 a _k is divergent.

Proof

Let’s prove that the series ∑ _k ≥ 1 a _k is absolutely convergent if l < 1 (the proof of its divergence in case l > 1 is completely analogous).

Letting l < 1, we can take a real number q such that l < q < 1. The convergence $\frac{\vert a_{n+1}\vert } {\vert a_{n}\vert } \stackrel{n}{\longrightarrow }l$ assures the existence of $n_{0} \in \mathbb{N}$ such that

$$\displaystyle{n \geq n_{0} \Rightarrow \frac{\vert a_{n+1}\vert } {\vert a_{n}\vert } \leq q.}$$

Hence, for n ≥ n ₀, we have

$$\displaystyle{\vert a_{n}\vert = \vert a_{n_{0}}\vert \prod _{k=n_{0}}^{n-1}\frac{\vert a_{k+1}\vert } {\vert a_{k}\vert } \leq \vert a_{n_{0}}\vert q^{n-n_{0} }.}$$

Thus, for n ≥ n ₀, the terms of the series ∑ _k ≥ 1 | a _k  | are majorized by those of the series $\sum _{k\geq 1}\vert a_{n_{0}}\vert q^{n-n_{0}}$, which converges, by Propositions 7.43 and 7.38. Therefore, it follows from the comparison test that ∑ _k ≥ 1 | a _k  | converges, which is the same as saying that ∑ _k ≥ 1 a _k is absolutely convergent. □ 

In the notations of the former proposition, we observe that, if l = 1, then the series ∑ _k ≥ 1 a _k may converge or diverge. Indeed, for $a_{n} = \frac{1} {n}$ we have

$$\displaystyle{\frac{a_{k+1}} {a_{k}} = \frac{k} {k + 1}\stackrel{k}{\longrightarrow }1,}$$

albeit the series $\sum _{k\geq 1} \frac{1} {k}$ diverges; on the other hand, for $a_{n} = \frac{1} {n^{2}}$ we have

$$\displaystyle{\frac{a_{k+1}} {a_{k}} = \frac{k^{2}} {(k + 1)^{2}}\stackrel{k}{\longrightarrow }1,}$$

while the series $\sum _{k\geq 1} \frac{1} {k^{2}}$ converges. On the positive side of things, we present the following

Example 7.52

Given a natural number m and a real number q > 1, explain whether the series $\sum _{k\geq 1}(-1)^{k-1}\frac{k^{m}} {q^{k}}$ converges or diverges.

Solution

Letting $a_{n} = (-1)^{n-1}\frac{n^{m}} {q^{n}}$ for n ≥ 1, we have

$$\displaystyle{\left \vert \frac{a_{n+1}} {a_{n}} \right \vert = \frac{(n + 1)^{m}} {q^{n+1}} \cdot \frac{q^{n}} {n^{m}} = \left (\frac{n + 1} {n} \right )^{\!\!m} \cdot \frac{1} {q}\stackrel{n}{\longrightarrow }\frac{1} {q} < 1.}$$

Therefore, by the ration test, the given series is absolutely convergent, hence, convergent. □ 

We close this section by discussing the product of two absolutely convergent series.

Theorem 7.53

Let ∑ _i≥1 a _i and ∑ _j≥1 b _j be absolutely convergent series. If

$$\displaystyle{ c_{k} =\sum _{i+j=k}a_{i}b_{j} =\sum _{ i=1}^{k-1}a_{ i}b_{k-i} }$$

(7.8)

for k ≥ 1, then ∑ _k≥1 c _k is absolutely convergent and such that

$$\displaystyle{\sum _{k\geq 1}c_{k} =\Big (\sum _{i\geq 1}a_{i}\Big)\Big(\sum _{j\geq 1}b_{j}\Big).}$$

Proof

It suffices to show that, given ε > 0, there exists $n_{0} \in \mathbb{N}$ such that, for n > n ₀, we have

$$\displaystyle{\Big\vert \sum _{k=1}^{2n}c_{ k} -\Big (\sum _{i\geq 1}a_{i}\Big)\Big(\sum _{j\geq 1}b_{j}\Big)\Big\vert <\epsilon \text{and} \Big\vert \sum _{k=1}^{2n-1}c_{ k} -\Big (\sum _{i\geq 1}a_{i}\Big)\Big(\sum _{j\geq 1}b_{j}\Big)\Big\vert <\epsilon.}$$

Let’s guarantee the existence of $n_{0} \in \mathbb{N}$ for which the first inequality above is true (the analysis of the validity of the second inequality is entirely analogous).

Given $n \in \mathbb{N}$, it follows from triangle inequality that

$$\displaystyle{ \begin{array}{rl} \Big\vert \sum _{k=1}^{2n}c_{k} -\Big (\sum _{i\geq 1}a_{i}\Big)\Big(\sum _{j\geq 1}b_{j}\Big)\Big\vert &\,\leq \Big\vert \sum _{k=1}^{2n}\Big(\sum _{i+j=k}a_{i}b_{j}\Big) -\Big (\sum _{i=1}^{n}a_{i}\Big)\Big(\sum _{j=1}^{n}b_{j}\Big)\Big\vert \\ &\quad +\Big \vert \Big(\sum _{i=1}^{n}a_{i}\Big)\Big(\sum _{j=1}^{n}b_{j}\Big) -\Big (\sum _{i\geq 1}a_{i}\Big)\Big(\sum _{j=1}^{n}b_{j}\Big)\Big\vert \\ &\quad +\Big \vert \Big(\sum _{i\geq 1}a_{i}\Big)\Big(\sum _{j=1}^{n}b_{j}\Big) -\Big (\sum _{i\geq 1}a_{i}\Big)\Big(\sum _{j\geq 1}b_{j}\Big)\Big\vert. \end{array} }$$

Let A, B and C denote the first, second and third summands of the right hand side above, respectively, so that

$$\displaystyle{B =\Big \vert \sum _{i>n}a_{i}\Big\vert \cdot \Big\vert \sum _{j=1}^{n}b_{ j}\Big\vert \ \ \text{and}\ \ C =\Big \vert \sum _{i\geq 1}a_{i}\Big\vert \cdot \Big\vert \sum _{j>n}b_{j}\Big\vert.}$$

The sequence $\left (\sum _{j=1}^{n}b_{j}\right )_{n\geq 1}$, being convergent, is bounded; therefore, there exists M > 0 such that $\Big\vert \sum _{j=1}^{n}b_{j}\Big\vert < M$, for every n ≥ 1. On the other hand, since the series ∑ _i ≥ 1 a _i and ∑ _j ≥ 1 b _j converge, we have

$$\displaystyle{\sum _{i>n}a_{i} =\sum _{i\geq 1}a_{i} -\sum _{i=1}^{n}a_{ i}\stackrel{n}{\longrightarrow }\sum _{i\geq 1}a_{i} -\sum _{i\geq 1}a_{i} = 0}$$

and, analogously, $\sum _{j>n}b_{j}\stackrel{n}{\longrightarrow }0$. In order to estimate C, we can suppose, without loss of generality, that ∑ _i ≥ 1 a _i ≠ 0. Then, we can choose $n_{1},n_{2} \in \mathbb{N}$ such that

$$\displaystyle{n > n_{1} \Rightarrow \Big\vert \sum _{i>n}a_{i}\Big\vert < \frac{\epsilon } {3M}\ \ \text{and}\ \ n > n_{2} \Rightarrow \Big\vert \sum _{j>n}b_{j}\Big\vert < \frac{\epsilon } {3\left \vert \sum _{i\geq 1}a_{i}\right \vert };}$$

hence, for n > max{n ₁, n ₂}, we have

$$\displaystyle{B < \frac{\epsilon } {3M} \cdot M = \frac{\epsilon } {3} \text{and} C <\Big \vert \sum _{i\geq 1}a_{i}\Big\vert \cdot \frac{\epsilon } {3\left \vert \sum _{i\geq 1}a_{i}\right \vert } = \frac{\epsilon } {3}.}$$

In what concerns A, notice firstly that

$$\displaystyle{ \begin{array}{rl} A&\, =\Big \vert \sum _{k=1}^{2n}\Big(\sum _{i+j=k}a_{i}b_{j}\Big) -\Big (\sum _{i=1}^{n}a_{i}\Big)\Big(\sum _{j=1}^{n}b_{j}\Big)\Big\vert =\Big \vert \sum _{\stackrel{\max \{i,j\}>n}{i+j\leq 2n}}a_{i}b_{j}\Big\vert \\ &\,\leq \sum _{\stackrel{\max \{i,j\}>n}{i+j\leq 2n}}\vert a_{i}b_{j}\vert \leq \sum _{i=n+1}^{2n}\sum _{j\leq n}\vert a_{i}\vert \vert b_{j}\vert +\sum _{ j=n+1}^{2n}\sum _{i\leq n}\vert a_{i}\vert \vert b_{j}\vert \\ &\,\leq \Big (\sum _{i=n+1}^{2n}\vert a_{i}\vert \Big)\Big(\sum _{j\geq 1}\vert b_{j}\vert \Big) +\Big (\sum _{j=n+1}^{2n}\vert b_{j}\vert \Big)\Big(\sum _{i\geq 1}\vert a_{i}\vert \Big). \end{array} }$$

To estimate A, we can suppose that ∑ _i ≥ 1 | a _i  | ≠ 0 and ∑ _j ≥ 1 | b _j  | ≠ 0. Now, since the series ∑ _i ≥ 1 | a _i  | and ∑ _j ≥ 1 | b _j  | converge, the sequences $\left (\sum _{i=1}^{n}\vert a_{i}\vert \right )_{n\geq 1}$ and $\left (\sum _{j=1}^{n}\vert b_{j}\vert \right )_{n\geq 1}$ are Cauchy; therefore, there exist $n_{3},n_{4} \in \mathbb{N}$ such that

$$\displaystyle{n > n_{3} \Rightarrow \sum _{i=n+1}^{2n}\vert a_{ i}\vert < \frac{\epsilon } {6\sum _{j\geq 1}\vert b_{j}\vert }\quad \text{and}\quad n > n_{4} \Rightarrow \sum _{j=n+1}^{2n}\vert b_{ j}\vert < \frac{\epsilon } {6\sum _{i\geq 1}\vert a_{i}\vert }.}$$

Then, for n > max{n ₃, n ₄}, we get

$$\displaystyle{A \leq \frac{\epsilon } {6\sum _{j\geq 1}\vert b_{j}\vert }\cdot \sum _{j\geq 1}\vert b_{j}\vert + \frac{\epsilon } {6\sum _{i\geq 1}\vert a_{i}\vert }\cdot \sum _{i\geq 1}\vert a_{i}\vert = \frac{\epsilon } {3}.}$$

Finally, by letting n ₀ = max{n ₁, n ₂, n ₃, n ₄} and taking n > n ₀, all of the previous estimates are valid, so that

$$\displaystyle{\Big\vert \sum _{k=1}^{2n}c_{ k} -\Big (\sum _{i\geq 1}a_{i}\Big)\Big(\sum _{j\geq 1}b_{j}\Big)\Big\vert \leq A + B + C < 3 \cdot \frac{\epsilon } {3} =\epsilon.}$$

 □ 

Problems: Section 7.4

• 1. Let (a _n ) _n ≥ 1 be a sequence of positive real numbers defined by $a_{1} = \frac{1} {2}$ and a _n+1 = a _n ² + a _n , for every $n \in \mathbb{N}$. Prove that $\sum _{k\geq 1} \frac{1} {a_{k}+1}$ converges and show that

  $$\displaystyle{\sum _{k\geq 1} \frac{1} {a_{k} + 1} = 2.}$$

• 2. Sequence (a _n ) _n ≥ 1 is a nonconstant AP of nonzero real numbers. Prove that $\sum _{k\geq 1} \frac{1} {a_{k}a_{k+1}}$ converges and compute its sum.

• 3. Given a real number a > 1, prove that the series $\sum _{k\geq 1}\frac{2k-1} {a^{k}}$ converges and compute its sum.

• 4. Decide whether the series $\sum _{k\geq 1} \frac{1} {\sqrt{k+\sqrt{k^{2 } -1}}}$ converges or diverges.

• 5. Prove that the series $\sum _{k>1000} \frac{1} {\sqrt{k^{3 } -1000k^{2}}}$ converges.

• 6. Let (a _n ) _n ≥ 1 be an infinite, nonconstant AP of positive terms. Prove that:

  • (a) $\sum _{k\geq 1} \frac{1} {a_{k}}$ diverges.

  • (b) $\sum _{k\geq 1} \frac{1} {a_{2^{k}}}$ converges.

• 7. (NMC) Let A be a finite set of naturals, all of the form 2^a3^b5^c, for some nonnegative integers a, b and c. Prove that

  $$\displaystyle{\sum _{x\in A}\frac{1} {x} < 4.}$$

• 8. * The purpose of this problem is to show that $e\cong 2.71828$, with five correct decimal places. To this end, do the following items:

  • (a) For every integer n > 10, show that

    $$\displaystyle{ \frac{1} {10!} + \frac{1} {11!} + \cdots + \frac{1} {n!} < \frac{1} {10!}\left (1 + \frac{1} {11} + \frac{1} {11^{2}} + \cdots + \frac{1} {11^{n-10}}\right ).}$$

  • (b) Use item (a), together with the fact that 10!  > 2 ⋅ 10⁶, to show that $0 < e -\sum _{k=1}^{10} \frac{1} {k!} < 10^{-6}$.

  • (c) Conclude from (b) that 2. 71828 approximates e with five correct decimal places.

• 9. * Given a sequence (a ₁, a ₂, a ₃, …) of digits, prove that there exists a single $x \in \mathbb{R}$ such that, for a given $n \in \mathbb{N}$, we have

  $$\displaystyle{0 \leq x -\left (\frac{a_{1}} {10} + \frac{a_{2}} {10^{2}} + \cdots + \frac{a_{k}} {10^{k}}\right ) \leq \frac{1} {10^{n}},}$$

  for every natural number k ≥ n. In such a case (and as the reader is certainly used to), we write x = 0. a ₁ a ₂ a ₃ … and say that 0. a ₁ a ₂ a ₃ … is the decimal representation of x.

• 10. Show that every real number x ∈ (0, 1) admits a unique decimal expansion of the form x = 0. a ₁ a ₂ a ₃ …, with a _n ≠ 0 for infinitely many values of n. Then, use this fact to construct a surjective function f: [0, 1] → [0, 1] × [0, 1].

• 11. Let (a _n ) _n ≥ 1 be a sequence of real numbers such that ∑ _k ≥ 1 a _k ² converges. Prove that, for every rational $\alpha > \frac{1} {2}$, $\sum _{k\geq 1}\frac{a_{k}} {k^{\alpha }}$ also converges.

• 12. Let (a _n ) _n ≥ 1 be a sequence of positive real numbers, such that the series ∑ _k ≥ 1 a _k converges. Prove that $\sum _{k\geq 1}\sqrt{a_{k } a_{k+1}}$ also converges.

• 13. Let (F _n ) _n ≥ 1 be the Fibonacci sequence , i.e., the sequence defined by F ₁ = 1, F ₂ = 1 and F _k+2 = F _k+1 + F _k , for every integer k ≥ 1. Show that the series $\sum _{k\geq 1} \frac{1} {F_{k}}$ converges.

• 14. Let (a _n ) _n ≥ 1 be a sequence of positive reals such that $\root{n}\of{a_{n}} \rightarrow l$.

  • (a) If l < 1, show that the series ∑ _k ≥ 1 a _k converges.

  • (b) If l > 1, show that the series ∑ _k ≥ 1 a _k diverges.

  • (c) If l = 1, give examples showing that the series ∑ _k ≥ 1 a _k may converge or diverge.

  The convergence criterion given by the case l < 1 is known as the root test .

• 15. Let ∑ _k ≥ 1 a _k be an absolutely convergent series, with ∑ _k ≥ 1 a _k  = 0. Show that

  $$\displaystyle{\sum _{n\geq 1}\left ( \frac{a_{1}} {(n - 1)^{2}} + \frac{a_{2}} {(n - 2)^{2}} + \cdots + \frac{a_{n-1}} {1^{2}} \right )}$$

  converges and compute the corresponding sum.

• 16. Prove that $e^{-1} =\sum _{k\geq 0}\frac{(-1)^{k}} {k!}$.

• 17. Let ∑ _k ≥ 1 a _k be an absolutely convergent series and $\varphi: \mathbb{N} \rightarrow \mathbb{N}$ be a bijection.

  • (a) If a _n  ≥ 0 for every $n \in \mathbb{N}$, prove that ∑ _k ≥ 1 a _k  = ∑ _k ≥ 1 a _φ(k).

  • (b) Write a _n  = a _n ⁺ − a _n ⁻, where a _n ⁺ = max{a _n , 0} and a _n ⁻ = −min{a _n , 0}, for every $n \in \mathbb{N}$. Prove that ∑ _k ≥ 1 a _k ⁺ and ∑ _k ≥ 1 a _k ⁻ are both convergent.

  • (c) Conclude that, in general, ∑ _k ≥ 1 a _k  = ∑ _k ≥ 1 a _φ(k).

  In view of item (c) above, we say that an absolutely convergent series is commutatively convergent .

  The convergence criterion for series stated in the next problem is due to N. Abel , and is known in the mathematical literature as Abel’s convergence test or Abel’s convergence criterion .

• 18. Let (a _n ) _n ≥ 1 and (b _n ) _n ≥ 1 be two sequences of real numbers satisfying the following conditions:

  • (a) The sequence (s _n ) _n ≥ 1, defined by s _n  = a ₁ + ⋯ + a _n for every $n \in \mathbb{N}$, is bounded.

  • (b) b ₁ ≥ b ₂ ≥ b ₃ ≥ ⋯ > 0 and $b_{n}\stackrel{n}{\longrightarrow }0$.

  Prove that the series ∑ _k ≥ 1 a _k b _k converges.

• 19. Show that Abel’s criterion implies Leibniz criterion.

• 20. Do the following items:

  • (a) Given $a,h \in \mathbb{R}$, with h ≠ 2l π for every $l \in \mathbb{Z}$, show that

    $$\displaystyle{\sum _{j=0}^{k}\sin (a + jh) = \frac{\sin \left (a + \frac{(k-1)h} {2} \right )\sin \frac{(k+1)h} {2} } {\sin \frac{h} {2} }.}$$

  • (b) Use Abel’s criterion to show that $\sum _{k\geq 1}\frac{\sin k} {k}$ converges.

• 21. In a cartesian coordinate system centered at O, let (A _n ) _n ≥ 1 be the sequence of points such that A ₁ = (1, 0) and:

  • (i) Triangle OA _n A _n+1 is rectangle at A _n and such that $\,\overline{A_{n}A_{n+1}} = 1$.

  • (ii) Triangles OA _n+1 A _n+2 and OA _n A _n+1 have disjoint interiors, for every n ≥ 1.

  Prove that, when n → +∞, half-line $\,\stackrel{\longrightarrow }{OA_{n}}$ revolves infinitely many times around O.

• 22. Let T _n be a right triangle whose side lengths are 4n ², 4n ⁴ − 1 and 4n ⁴ + 1, with $n \in \mathbb{N}$. Let α _n be the measure, in radians, of its internal angle opposite to the side of length 4n ². Show that

  $$\displaystyle{\sum _{k\geq 1}\alpha _{k} = \frac{\pi } {2}.}$$