An Introduction to Infinite Series

• Daniel Rosenthal
• David Rosenthal
• Peter Rosenthal
Chapter
Part of the Undergraduate Texts in Mathematics book series (UTM)

Abstract

An infinite series is an expression of the form a1 + a2 + a3 + ⋯, where each ai is a real number. We discuss the question of when an infinite series has a “sum” in a precise sense that we will explain. A series that does have such a sum is said to “converge.” The basic properties of convergence of infinite series are rigorously proven. In particular, it is shown that every “infinite decimal” converges.

What is .3333… (where the sequence of 3’s continues forever)? Presumably, this expression means
\displaystyle \begin{aligned}\frac{3}{10} + \frac{3}{10^2} + \frac{3}{10^3} + \frac{3}{10^4} + \cdots\end{aligned}
where the “⋯” indicates that the “sum” continues indefinitely. But then, what does that mean? What does it mean to add up an infinite number of terms? Is the sum $$\frac {1}{3}$$? Or is it merely close to $$\frac {1}{3}$$? Does it have a precise meaning?
Similarly, what is
\displaystyle \begin{aligned}\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots\end{aligned}
(where the indicated sum continues forever)? Expressions such as these are called “infinite series.” In some cases, as we shall see, there is a natural way of defining the sum of an infinite series.

In this chapter, we present the basic facts about infinite series, emphasizing an understanding of the fundamental concepts. In order to make the central idea more accessible, our approach is a little unorthodox; the more standard approach is explained in Problem 13.8 at the end of the chapter.

13.1 Convergence

The rough idea is to define the sum of an infinite series to be S if adding any large enough (but finite) number of terms produces a sum that is very close to S. We will have to make precise what is meant by “very close.” Before we present the formal definitions, we discuss the following example:
\displaystyle \begin{aligned}\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots\end{aligned}
We know what is meant by $$\frac {1}{2} + \frac {1}{4}$$; it is $$\frac {3}{4}$$. We know what is meant by $$\frac {1}{2} + \frac {1}{4} + \frac {1}{8}$$; it is $$\frac {7}{8}$$. Similarly, we know what is meant by $$\frac {1}{2} + \frac {1}{4} + \frac {1}{8} + \frac {1}{16}$$, or by the sum of any finite number of terms of this series. The sum of all the infinitely many terms is defined as a kind of a limit, as we now explain.
For each natural number n, let $$\textit {S}_n = \frac {1}{2} + \frac {1}{4} + \frac {1}{8} + \cdots + \frac {1}{2^n}$$ denote the sum of the first n terms of the series. We seek a simpler formula for Sn. Multiplying both sides of the equation defining Sn by $$\frac {1}{2}$$ yields $$\frac {1}{2} \textit {S}_n = \frac {1}{4} + \frac {1}{8} + \frac {1}{16} + \cdots + \frac {1}{2^{n+1}}$$. Therefore,
\displaystyle \begin{aligned} \textit{S}_n - \frac{1}{2}\textit{S}_n &= \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots + \frac{1}{2^n}\right) - \left(\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots + \frac{1}{2^{n+1}}\right) \\ &=\frac{1}{2} - \frac{1}{2^{n+1}}\end{aligned}
In other words, $$\frac {1}{2} \textit {S}_n = \frac {1}{2} - \frac {1}{2^{n+1}}$$. Thus, $$\textit {S}_n = 2\big (\frac {1}{2} - \frac {1}{2^{n+1}}\big ) = 1 - \frac {1}{2^n}$$.

This formula shows that Sn is close to 1 if n is large. In fact, no matter how close we want Sn to be to 1, we can guarantee that Sn will be at least that close by choosing n sufficiently large. To see this, note that the difference between Sn and 1 is $$\frac {1}{2^n}$$. Thus, for example, we can guarantee that Sn is within $$\frac {1}{10}$$ of 1 by taking n to be greater than or equal to 4, since $$\frac {1}{16}$$ is less than $$\frac {1}{10}$$. We can guarantee that Sn is within $$\frac {1}{1000}$$ of 1 by choosing n to be greater than or equal to 10, since $$\frac {1}{2^{10}}$$ (that is, $$\frac {1}{1024}$$) is less than $$\frac {1}{1000}$$. Since Sn is very close to 1 when n is large, the definition of the sum of an infinite series that we give below (Definition 13.1.4) implies that the sum of this infinite series is 1.

To define precisely what we mean by the sum of an infinite series, it is useful to have a notation for the distance between two real numbers. The following definition will play a central role.

Definition 13.1.1

The absolute value of a real number a, denoted |a|, is defined to be the distance from a to 0 on the number line. More precisely, if a is positive or zero, then |a| is just a, and if a is negative, then |a| is equal to − a.

For example, |1| = 1, |− 5| = −(−5) = 5, $$\left |-\frac {3}{\sqrt {2}}\right | = -\left (-\frac {3}{\sqrt {2}}\right )=\frac {3}{\sqrt {2}}$$, and |0| = 0. Therefore, the absolute value has the effect of “making a number nonnegative.”

Consider the question: “Which real numbers a satisfy |a| = 5?” That is, “Which real numbers a have a distance of 5 from 0?” As can been seen by looking at the real line (as in Figure 13.1), there are only two directions one can move away from zero: the positive direction and the negative direction. Thus, there are only two real numbers that have a distance of 5 from 0; namely, 5 and − 5. More generally, for every positive real number d, there are precisely two real numbers with absolute value equal to d; namely, d and − d.

What does |a|≤ 5 mean? This means that a has a distance from 0 that is less than or equal to 5. Looking at the real number line, we see that this means that a cannot be larger than 5 and also cannot be less than − 5. In other words, |a|≤ 5 is equivalent to − 5 ≤ a ≤ 5. More generally, if d > 0, then |a|≤ d is equivalent to − d ≤ a ≤ d.

The distance between two numbers a and b can be expressed using absolute values as |a − b|. We show this as follows. Since this is clearly true when a or b is zero, we need only consider the cases where a and b are both nonzero. When a and b are both positive numbers it is not hard to see that the distance between them is |a − b|; simply picture a and b on a number line, and observe that, regardless of which is larger, the larger minus the smaller is |a − b| (which is the same as |b − a|). For example, the distance between 2 and 5 is 3, and whether we write |5 − 2| or |2 − 5|, we get 3. The case where a and b are both negative numbers can be obtained from the previous case by noting that the distance between a and b is the same as the distance between − a and − b, which is |(−a) − (−b)| = |− a + b| = |− (a − b)| = |a − b|.

The case that takes a little more thought is the case where one of a and b is positive and the other is negative. Suppose that a is negative and b is positive (the other way around is handled analogously). Looking at the number line (Figure 13.2), since 0 is between a and b, we see that the distance between a and b is equal to the distance from a to 0 plus the distance from b to 0. That is, the distance between a and b is equal to |a| + |b|. Since a < 0 and b > 0, |a| + |b| is equal to (−a) + b = b − a = |b − a| = |a − b|. Thus, the distance between a and b is |a − b| in all cases.

Using absolute values to denote distances between numbers is useful in formulating the precise definition of what is meant by the sum of an infinite series.

We return to the example $$\frac {1}{2} + \frac {1}{4} + \frac {1}{8} + \cdots$$. We want to capture the idea that, no matter how close we require the sum $$\textit {S}_n = \frac {1}{2} + \frac {1}{4} + \frac {1}{8} + \cdots + \frac {1}{2^n}$$ to be to 1, we can get it that close by including a sufficient number of terms; i.e., by choosing n to be sufficiently large.

It is common in mathematics to use the Greek letter epsilon, written in the form ε, to denote a small positive real number, as in the following. For every positive number ε, there is a natural number N such that Sn is within ε of 1 whenever n is greater than or equal to N. This can be reformulated in terms of absolute values as the statement: For every ε > 0 there is a natural number N such that |Sn − 1| < ε whenever n ≥ N. This precisely captures what is meant by the requirement that Sn gets arbitrarily close to 1 when n is sufficiently large.

To see that this precise statement holds for $$\textit {S}_n = \frac {1}{2} + \frac {1}{4} + \frac {1}{8} + \cdots + \frac {1}{2^n} = 1 - \frac {1}{2^n}$$, first note that $$|\textit {S}_n - 1| = \big |-\frac {1}{2^n}\big | = \frac {1}{2^n}$$. Therefore, we must show that, for every ε > 0, there exists a natural number N such that $$\frac {1}{2^n} < \varepsilon$$ whenever n ≥ N.

Let ε be any positive number. We find such an N by first noting that, since powers of 2 can be arbitrarily large, there is an N such that $$2^N > \frac {1}{\varepsilon }$$. Fix such an N. Multiplying both sides of $$\frac {1}{\varepsilon } < 2^N$$ by $$\frac {\varepsilon }{2^N}$$ gives $$\frac {1}{2^N} < \varepsilon$$. If n ≥ N, then $$\frac {1}{2^n} \leq \frac {1}{2^N} < \varepsilon$$. Therefore, for each ε > 0, if N is large enough that 2N is greater than $$\frac {1}{\varepsilon }$$, then Sn is within ε of 1 whenever n ≥ N, as desired. This proves that, according to the general definition we give below (13.1.4), the sum of the infinite series $$\frac {1}{2} + \frac {1}{4} + \frac {1}{8} + \frac {1}{16} + \cdots$$ is 1.

Definition 13.1.2

An infinite series is an expression of the form
\displaystyle \begin{aligned}a_1 + a_2 + a_3 + \cdots\end{aligned}
where the ai are real numbers and the indicated sum continues forever. The ai are called the terms of the series.
Some examples of infinite series are:
\displaystyle \begin{aligned}1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots\end{aligned}
\displaystyle \begin{aligned}1 + \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \cdots\end{aligned}
\displaystyle \begin{aligned}1 + \frac{1}{\sqrt 2} + \frac{1}{\sqrt 3} + \frac{1}{\sqrt 4} + \cdots\end{aligned}
\displaystyle \begin{aligned}1 - 1 + 1 -1 + \cdots\end{aligned}
\displaystyle \begin{aligned}79^2 + 80^2 + 81^2 + 82^2 + \cdots\end{aligned}
\displaystyle \begin{aligned}\frac{2}{3} + \frac{2^2}{3^2} + \frac{2^3}{3^3} + \cdots\end{aligned}
\displaystyle \begin{aligned}\frac{4}{12^3} + \frac{4}{13^3} + \frac{4}{14^3} + \frac{4}{15^3} + \cdots\end{aligned}

Any particular infinite series may or may not have a sum according to the definition given below. The general definition is the same as the precise formulation in the special case considered above where $$\textit {S}_n = 1 - \frac {1}{2^n}$$. The definition captures the rough idea that adding a large but finite number of terms of the series gives a partial sum that is close to the “sum” of the entire series.

Definition 13.1.3

For an infinite series
\displaystyle \begin{aligned}a_1 + a_2 + a_3 + a_4 + \cdots \end{aligned}
the nth partial sum , often denoted by Sn, is the sum of the first n terms. That is,
\displaystyle \begin{aligned}\textit{S}_n = a_1 + a_2 + a_3 + \cdots + a_n\end{aligned}

Definition 13.1.4

The infinite series with partial sums Sn converges to S (or has sum S ) if, for every ε > 0, there is a natural number N such that |Sn −S| < ε whenever n ≥ N.

As we showed above, the series $$\frac {1}{2} + \frac {1}{4} + \frac {1}{8} + \cdots$$ converges to 1 according to this definition.

Note that if there is any N that satisfies the definition for a particular ε, then there are infinitely many such N, for if N0 satisfies the definition, so does any N larger than N0.

Note also that a series does not converge to a given S if there is any ε greater than 0 for which there is no N satisfying the definition. (Of course, if ε0 is such an ε, then so is any smaller positive number.)

For some infinite series, such as the example above where S = 1, it is not hard to determine the sum. There are other infinite series that can be shown to converge but for which finding an expression for the sum (other than the infinite series itself) is very difficult or even impossible. Moreover, there are infinite series that do not converge to any sum at all.

Definition 13.1.5

An infinite series converges if there is some S for which the series converges to S according to Definition 13.1.4. If a series does not converge to any S, we say that the series diverges .

Example 13.1.6

The series 1 + 1 + 1 + ⋯ diverges.

Proof

The partial sums of this series are: S1 = 1, S2 = 2, S3 = 3, and, for each natural number n, Sn = n. We want to show that there is no S that satisfies the definition of sum (13.1.4) for this series. Let S be any real number. If n is sufficiently large, then Sn will be much larger than S. To show that S is not the sum of the series, it suffices to find any ε > 0 for which there is no N satisfying Definition 13.1.4. For this particular series, in fact, every ε > 0 has that property. For instance, take ε = 3. No matter what N is chosen, there are an infinite number of n’s greater than N for which Sn is greater than S + 3. For those Sn, it is not the case that |Sn − S| < 3. Therefore, the series diverges. □

Example 13.1.7

The series 1 − 1 + 1 − 1 + 1 − 1 + ⋯ diverges.

Proof

Consider the partial sums of this series: S1 = 1, S2 = 0, S3 = 1, and so on. That is, the odd partial sums are all 1 and the even partial sums are all 0. To show that there is no S to which the series converges, it suffices to find some ε > 0 for which there is no N satisfying Definition 13.1.4. In this example, $$\varepsilon = \frac {1}{3}$$ (for instance) has that property. For, no matter what N is chosen, there will exist an n1 > N (any odd n1 > N) and an n2 > N (any even n2 > N) such that $$\textit {S}_{n_1} = 1$$ and $$\textit {S}_{n_2} = 0$$. There is no real number S that is within $$\frac {1}{3}$$ of both 1 and 0. Thus, the series diverges. □

The proofs of many results concerning infinite series require an important inequality concerning absolute values. It is called the “triangle inequality” because its generalization to vectors in the plane is equivalent to the fact that the sum of the lengths of two sides of a triangle is greater than or equal to the length of the third side (see Theorem in Chapter ).

The Triangle Inequality 13.1.8

Let x and y be real numbers. Then
\displaystyle \begin{aligned}|x+y|\leq |x|+|y|\end{aligned}

Proof

Recall from our discussion about absolute values that if d > 0, then |a|≤ d is equivalent to − d ≤ a ≤ d. Therefore, we can prove the Triangle Inequality by showing that − (|x| + |y|) ≤ x + y ≤|x| + |y|. For this, observe that for every real number a, −|a|≤ a ≤|a|. Thus, in particular, −|x|≤ x ≤|x| and −|y|≤ y ≤|y|. Adding these two inequalities gives us the desired inequality, − (|x| + |y|) ≤ x + y ≤|x| + |y|. □

A fundamental question is: can a series have two different sums? That is, can a series converge to S and also converge to T, with S different from T? Our first application of the triangle inequality is in providing an answer to this question.

Theorem 13.1.9

An infinite series converges to at most one real number. That is, if a series converges to S and also converges to T, then S = T.

Proof

Let Sn be the nth partial sum of a given series, and suppose that the series converges to S and also to T. We will show that, for every ε > 0, |S − T| < ε. Since 0 is the only nonnegative real number that is less than every positive real number, it will then follow that |S − T| = 0; that is, S = T.

Let ε > 0 be given. For every n, the Triangle Inequality (13.1.8) implies
\displaystyle \begin{aligned}|\textit{S}-T| = |\textit{S} - \textit{S}_n + \textit{S}_n - T| = |(\textit{S} - \textit{S}_n) + (\textit{S}_n -\textit{T})| \leq |\textit{S} - \textit{S}_n| + |\textit{S}_n - T|\end{aligned}
Since the series converges to S, for every ε1 > 0 there is an N1 such that |Sn −S| < ε1 for every n ≥ N1. Similarly, since the series converges to T, for every ε2 > 0 there is an N2 such that |Sn − T| < ε2 for every n ≥ N2. Choose $$\varepsilon _1 = \frac {\varepsilon }{2}$$ and $$\varepsilon _2 = \frac {\varepsilon }{2}$$. If N is the larger of N1 and N2, then both inequalities are satisfied for all n ≥ N. Thus, for n ≥ N,
\displaystyle \begin{aligned}|\textit{S} - \textit{S}_n| + |\textit{S}_n - T| < \varepsilon_1 + \varepsilon_2 = \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon\end{aligned}
Therefore, |S − T| < ε, as desired. □

Despite the previous theorem, rearranging the order of the terms of an infinite series sometimes produces a series with a different sum (see Problem 13.8). That is, the order in which the terms of an infinite series are added is important under certain conditions (see Problems 13.8 and 13.8). This surprising possibility is, of course, different from the situation when adding a finite number of numbers. It is therefore important to note that the definition of a particular infinite series depends not only on the terms of the series but also on the order in which the terms are arranged. However, if the terms of the series are all nonnegative, then rearranging the order of the terms does not affect the sum (see Problem 13.8).

13.2 Geometric Series

The example $$\frac {1}{2} + \frac {1}{4} + \frac {1}{8} + \cdots$$ that we considered above is a particular instance of a special kind of series that is easily dealt with.

Definition 13.2.1

A series a1 + a2 + a3⋯ is called a geometric series if there is a number r, called the ratio of the series, such that each term of the series is obtained from the preceding term by multiplying by r. That is, there is a number r such that, if the first term of the series is a, then the second term is ar, the third term is ar2, and so on. Such a series has the form
\displaystyle \begin{aligned}a + ar^2 + ar^3 + \cdots + ar^n + \cdots\end{aligned}
for some real numbers a and r.
The series that we previously discussed,
\displaystyle \begin{aligned}\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots\end{aligned}
is a geometric series with first term $$a = \frac {1}{2}$$ and ratio $$r = \frac {1}{2}$$.

We begin the analysis of geometric series by finding a formula for the nth partial sum of a general geometric series.

Theorem 13.2.2

If a is a real number and r is a real number other than 1, and if
\displaystyle \begin{aligned}\textit{S}_n = a + ar+ ar^2 + ar^3 + \cdots + ar^{n-1}\end{aligned}
then
\displaystyle \begin{aligned}\textit{S}_n = \frac{a - ar^{n}}{1-r}\end{aligned}

Proof

The proof is similar to that of the special case that we discussed above (where $$a = \frac {1}{2}$$ and $$r = \frac {1}{2}$$). Since
\displaystyle \begin{aligned}r\textit{S}_n = ar + ar^2 + \cdots + ar^n\end{aligned}
it follows that Sn − rSn = a − arn (since all of the other terms cancel each other out when doing the subtraction). Thus, (1 − r)Sn = a − arn, or $$\textit {S}_n = \frac {a - ar^n}{1 - r}$$. (Note that this formula for Sn does not make sense when r = 1. Of course, Sn is simply na in this case.) □

We will use the above formula to determine which geometric series converge and to find the sum of a geometric series when it exists. We need the fact that if a number has absolute value less than 1, then its powers get arbitrarily small. That is, we need the following lemma. (Like ε, the Greek letter δ, read “delta,” is often used to denote a small positive number.)

Lemma 13.2.3

If |r| < 1, then for every positive number δ there is a natural number N such that |r|n is less than δ for all n  N.

Proof

If r = 0, then rn = 0 for all n, and therefore any N will do. If r is not 0, then |r| < 1 implies that $$\frac {1}{|r|}$$ is greater than 1. Define the positive number t by $$t = \frac {1}{|r|} - 1$$, so $$\frac {1}{|r|} = 1 + t$$. We need the fact that, for every natural number n, (1 + t)n is greater than or equal to 1 + nt (which can be proven easily by mathematical induction, as stated in Problem 5 of this chapter). Now let δ > 0 be given. Choose any N that is large enough so that Nt is greater than $$\frac {1}{\delta }$$. Suppose that n ≥ N. Then $$nt \geq Nt > \frac {1}{\delta }$$. Thus, $$(1 + t)^n \geq (1 + nt) > 1 + \frac {1}{\delta } > \frac {1}{\delta }$$, so $$(1 + t)^n > \frac {1}{\delta }$$, or $$\left (\frac {1}{|r|}\right )^n=\frac {1}{|r|{ }^n} > \frac {1}{\delta }$$. This implies that |r|n < δ, as desired. □

Theorem 13.2.4

If a is a real number and r is a real number with |r| < 1, then the geometric series a + ar + ar2 + ⋯ + arn−1 + ⋯ converges to $$\frac {a}{1- r}$$.

Proof

The nth partial sum of the series is $$\textit {S}_n = \frac {a - ar^n}{1 - r}$$ (Theorem 13.2.2). According to the precise definition of convergence (13.1.4), the theorem will be proven if we establish that for every ε > 0 there exists an N such that $$\big |\textit {S}_n - \frac {a}{1-r} \big | < \varepsilon$$ whenever n ≥ N. Note that the difference between $$\frac {a}{1-r}$$ and Sn is
\displaystyle \begin{aligned}\frac{a}{1-r} - \frac{a - ar^n}{1-r} = \frac{ar^n}{1-r}\end{aligned}
Let ε > 0 be given. If a = 0, then $$\frac {ar^n}{1-r}=0$$. Assume now that a ≠ 0. By Lemma 13.2.3, there is an N such that for all n ≥ N, |r|n is less than the positive number $$\frac {1 - r}{|a|}\cdot \varepsilon$$. Then, for all n ≥ N,
\displaystyle \begin{aligned}\left|\textit{S}_n - \frac{a}{1-r} \right|=\left|\frac{ar^n}{1-r}\right| = \frac{|a|}{1-r}\cdot |r^n| < \frac{|a|}{1-r} \cdot \frac{1-r}{|a|}\cdot \varepsilon = \varepsilon\end{aligned}
Therefore, $$\frac {a}{1-r}$$ is the sum of the series. □

The example with which we began is a special case of the above.

Example 13.2.5

$$\frac {1}{2} + \frac {1}{4} + \frac {1}{8} + \cdots = 1$$

Proof

This is the special case of Theorem 13.2.4 where $$a = \frac {1}{2}$$ and $$r = \frac {1}{2}$$. □

There are, of course, many other geometric series.

Example 13.2.6

$$\sqrt {17} + \frac {\sqrt {17}}{3} + \frac {\sqrt {17}}{3^2} + \frac {\sqrt {17}}{3^3} + \cdots = \frac {3\sqrt {17}}{2}$$

Proof

This follows from Theorem 13.2.4 using $$a = \sqrt {17}$$ and $$r = \frac {1}{3}$$. The sum is $$\frac {a}{1-r} = \frac {\sqrt {17}}{\left (\frac {2}{3}\right )} = \frac {3 \sqrt {17}}{2}$$. □

Geometric series can have negative ratios.

Example 13.2.7

$$1 - \frac {1}{2} + \frac {1}{4} - \frac {1}{8} + \frac {1}{16} - \cdots = \frac {2}{3}$$

Proof

This is the case of Theorem 13.2.4 where a = 1 and $$r = -\frac {1}{2}$$. The sum is $$S = \frac {1}{\left (1 - (-\frac {1}{2})\right )} = \frac {1}{\left (\frac {3}{2}\right )} = \frac {2}{3}$$. □

The proper interpretation of .333333… is as an infinite series; it is another way of writing the series $$\frac {3}{10} + \frac {3}{100} + \frac {3}{1000} + \cdots$$. Hence, we can determine its value.

Example 13.2.8

The infinite decimal .33333… is $$\frac {1}{3}$$.

Proof

This follows from Theorem 13.2.4 with $$a = \frac {3}{10}$$ and $$r = \frac {1}{10}$$; the sum is $$\frac {\big (\frac {3}{10}\big )}{\big (1 - \frac {1}{10}\big )} = \frac {\big (\frac {3}{10}\big )}{\big (\frac {9}{10}\big )} = \frac {1}{3}$$. □

We shall see in Section 13.6 that every infinite decimal .b1b2b3…, where each bi is an integer between 0 and 9, is a convergent series (Theorem 13.6.2), although most infinite decimals are not geometric series.

13.3 Convergence of Related Series

In some cases the convergence of a series can be established by using the fact that a related series is known to converge.

Theorem 13.3.1

If a1 + a2 + a3 + ⋯ is an infinite series that converges to S and c is any real number, then the infinite series ca1 + ca2 + ca3 + ⋯ (obtained by multiplying each of the terms of the original series by c) converges to cS.

Proof

We want to show that the sum of the series ca1 + ca2 + ca3 + ⋯ is cS. By Definition 13.1.4, we need to show that, for every ε > 0, there exists a natural number N such that the absolute value of the difference between cS and the nth partial sum of the series is less than ε for every n ≥ N. In the case where c = 0, the series obviously converges to 0 ⋅S = 0. We therefore assume that c ≠ 0 in what follows.

Let Sn be the nth partial sum of the series a1 + a2 + a3 + ⋯. Then the nth partial sum of the series ca1 + ca2 + ca3 + ⋯ is cSn. Therefore, we need to consider |cSn − cS|. But this is equal to |c(Sn −S)| = |c|⋅|Sn −S|. This enables us to prove the theorem, as follows. Let ε > 0 be given. Since a1 + a2 + a3 + ⋯ converges to S, there is a natural number N such that Sn is within the positive number $$\frac {\varepsilon }{|c|}$$ of S for every n ≥ N. That is, $$|\textit {S}_n-\textit {S}|<\frac {\varepsilon }{|c|}$$ for every n ≥ N. Therefore, $$|c\textit {S}_n-c\textit {S}|=|c|\cdot |\textit {S}_n-\textit {S}|<|c|\cdot \frac {\varepsilon }{|c|}=\varepsilon$$ for every n ≥ N, so Definition 13.1.4 is satisfied by cS. □

Example 13.3.2

The infinite decimal .66666… is equal to $$\frac {2}{3}$$ and the infinite decimal .99999… is equal to 1.

Proof

We could establish both of these facts using Theorem 13.2.4, as we did in Example 13.2.8. But, since we already know that the infinite decimal .33333… is $$\frac {1}{3}$$, we can use the theorem we just proved (13.3.1) to get the result even more simply. The infinite decimal .66666… is twice the infinite decimal .33333…. More precisely, each of the terms of the infinite series representing .66666… is equal to twice the corresponding term of the infinite series that represents .33333…. Thus, by Theorem 13.3.1, .66666… is equal to $$2\cdot \frac {1}{3}=\frac {2}{3}$$. Similarly, since .99999… is 3 times .33333…, Theorem 13.3.1 implies that .99999… is $$3\cdot \frac {1}{3}=1$$. □

Theorem 13.3.3

If a1 + a2 + a3 + ⋯ is an infinite series that converges to S and b1 + b2 + b3 + ⋯ is an infinite series that converges to T, then the infinite series (a1 + b1) + (a2 + b2) + (a3 + b3) + ⋯ converges to S + T.

Proof

We must show that the sum of the series (a1 + b1) + (a2 + b2) + (a3 + b3) + ⋯ is S + T. By Definition 13.1.4, we need to show that, for every ε > 0, there exists a natural number N such that the absolute value of the difference between S + T and the nth partial sum of the series is less than ε for every n ≥ N.

Let Sn be the nth partial sum of the series a1 + a2 + a3 + ⋯, and let Tn be the nth partial sum of the series b1 + b2 + b3 + ⋯. Then Sn + Tn is the nth partial sum of the series (a1 + b1) + (a2 + b2) + (a3 + b3) + ⋯. Therefore, we need to analyze |(Sn + Tn) − (S + T)|. First, |(Sn + Tn) − (S + T)| = |(Sn −S) + (Tn − T)|. By the Triangle Inequality (13.1.8), |(Sn −S) + (Tn − T)|≤|Sn −S| + |Tn − T|. Thus, |(Sn + Tn) − (S + T)|≤|Sn −S| + |Tn − T|.

This inequality enables us to prove the theorem, as follows. Let ε > 0 be given. Since a1 + a2 + a3 + ⋯ converges to S, there exists an integer N1 such that $$|\textit {S}_n-S|<\frac {\varepsilon }{2}$$ for every n ≥ N1. Since b1 + b2 + b3 + ⋯ converges to T, there is an integer N2 such that $$|\textit {T}_n-T|<\frac {\varepsilon }{2}$$ for every n ≥ N2. Thus, if we let N be the larger of N1 and N2, then both of these inequalities hold whenever n ≥ N. Therefore, $$|(\textit {S}_n+\textit {T}_n)-(\textit {S}+T)| \leq |\textit {S}_n-\textit {S}|+|\textit {T}_n-T|<\frac {\varepsilon }{2}+\frac {\varepsilon }{2}=\varepsilon$$ for all n ≥ N. □

The above theorem can be summarized “convergent series can be added term-by-term.”

13.4 Least Upper Bounds

In order to investigate infinite series other than geometric series, we need to understand an important property of the real numbers called “The Least Upper Bound Property.”

Definition 13.4.1

If $$\mathcal {S}$$ is a set of real numbers, then the real number t is an upper bound for $$\mathcal {S}$$ if t is greater than or equal to every element of $$\mathcal {S}$$. That is, t is an upper bound of $$\mathcal {S}$$ if x ≤ t for every x in $$\mathcal {S}$$.

For example, 2 is an upper bound of the set $$\big \{1, \frac {1}{2}, \frac {1}{3}, \dots \big \}$$; − 1 is an upper bound of {−1, −2, −3, … }; 5 is an upper bound of $$\big \{3 - \frac {1}{2}, 3 - \frac {1}{3}, 3 - \frac {1}{4}, \dots \big \}$$; and 28 is an upper bound of $$\{x : x < \sqrt {2}\}$$.

It is important to observe that if t is an upper bound of a set $$\mathcal {S}$$, then every real number greater than t is also an upper bound of $$\mathcal {S}$$. Thus, if a set $$\mathcal {S}$$ has an upper bound, it has infinitely many upper bounds. Some sets, however, such as {1, 2, 3, 4, 5, … } and $$\Big \{\big (\sqrt {2}\big )^n : n \in \mathbb {N}\Big \}$$, do not have any upper bounds at all.

Definition 13.4.2

If $$\mathcal {S}$$ is a set of real numbers, then the real number t is a least upper bound of $$\mathcal {S}$$ if t is an upper bound of $$\mathcal {S}$$ and every upper bound of $$\mathcal {S}$$ is greater than or equal to t. That is, a least upper bound is a smallest upper bound.

For example, a least upper bound of $$\big \{1, \frac {1}{2}, \frac {1}{3}, \dots \big \}$$ is 1. This can be seen as follows. Since every number in the set is less than or equal to 1, 1 is an upper bound. If t is any upper bound, then t is greater than or equal to 1 (since 1 is in the set), so 1 is a least upper bound.

A least upper bound of {−1, −2, −3, −4, −5, … } is − 1. To see this, first observe that every number in the set is less than or equal to − 1, so − 1 is an upper bound. Every upper bound must be greater than or equal to − 1 since − 1 is in the set.

A least upper bound of $$\big \{3 - \frac {1}{2}, 3 - \frac {1}{3}, 3 - \frac {1}{4}, \dots \big \}$$ is 3. To verify this, begin by noting that, since 3 is greater than any number in the set, 3 is an upper bound. To see that it is a least upper bound, note that if t is any upper bound, then t must be greater than or equal to $$3 - \frac {1}{n}$$ for every natural number n. If t was less than 3, then there would be some n such that $$t < 3 - \frac {1}{n}$$ (just choose n such that $$\frac {1}{n} < 3 - t$$). Thus, t would not be an upper bound. Therefore, all upper bounds are greater than or equal to 3, so 3 is a least upper bound. Note that, in this example, 3 is a least upper bound of the set but is not in the set.

A least upper bound of $$\big \{x : x < \sqrt {2}\big \}$$ is $$\sqrt {2}$$. To see this, note that every number in the set is less than $$\sqrt {2}$$, so $$\sqrt {2}$$ is an upper bound. If t is a number less than $$\sqrt {2}$$, then $$t + \frac {1}{2} \big (\sqrt {2} - t\big ) = \frac {1}{2} (t + \sqrt {2}) < \frac {1}{2} (\sqrt {2} + \sqrt {2}) = \sqrt {2}$$. Therefore $$t + \frac {1}{2} \big (\sqrt {2} - t\big )$$ is in the set. But $$t + \frac {1}{2} \big (\sqrt {2} - t\big )$$ is obviously greater than t, since $$\sqrt {2} - t$$ is positive. Therefore such a t is not an upper bound for the set. Thus, $$\sqrt {2}$$ is a least upper bound.

Suppose that $$\mathcal {S}$$ is the empty set; that is, the set that does not contain any elements. Then every real number is an upper bound for $$\mathcal {S}$$, for, no matter what real number is given, $$\mathcal {S}$$ does not contain any numbers greater than it. Since every real number is an upper bound for the empty set, the empty set does not have a least upper bound.

A crucial property of the real numbers that we will need in order to understand infinite series, and which is also important in many other contexts, is the existence of least upper bounds. We assume as an axiom that the real numbers have this property. (In fact, this property can be proven from the Dedekind cuts construction of the real numbers in terms of sets of rational numbers; see Problem 13.8 at the end of this chapter.)

The Least Upper Bound Property 13.4.3

Every nonempty set of real numbers that has an upper bound has a least upper bound. In other words, the set of upper bounds has a smallest element. More precisely, if $$\mathcal {S}$$ is any set other than the empty set and $$\mathcal {S}$$ has an upper bound, then there is an upper bound t0 for $$\mathcal {S}$$ such that every upper bound for $$\mathcal {S}$$ is greater than or equal to t0.

We next show that a set cannot have two different least upper bounds.

Theorem 13.4.4

If a nonempty set of real numbers has an upper bound, then the set has a unique least upper bound.

Proof

By the Least Upper Bound Property (13.4.3), every nonempty set of real numbers that has an upper bound has a least upper bound; we must show that there is at most one least upper bound for a given set. Suppose that both t1 and t2 are least upper bounds of the same nonempty set. Then, since t1 is a least upper bound and t2 is another upper bound, t1 ≤ t2. Similarly, since t2 is a least upper bound and t1 is another upper bound, t2 ≤ t1. Therefore, t1 = t2. □

Theorem 13.4.5

If an infinite series converges, then the set of partial sums of the series has an upper bound.

Proof

Suppose that an infinite series converges to the sum S. Then, by Definition 13.1.4, for every ε > 0 there exists an N such that |Sn −S| < ε whenever n ≥ N. In particular, there is such an N0 for ε = 1. That is, the absolute value of Sn −S is less than 1 whenever n is greater than or equal to N0. It follows, in particular, that for all n ≥ N0, Sn −S < 1. Therefore, Sn < 1 + S for every n ≥ N0. Thus, we have found an upper bound, 1 + S, for the set of all partial sums except for the first N0 − 1 of them. Let t be the largest of the numbers in the finite set $$\{\textit {S}_1, \textit {S}_2, \dots , \textit {S}_{N_0}\}$$. Then the larger of 1 + S and t is an upper bound for the set of all partial sums. □

The converse of Theorem 13.4.5 is not true in general, as the next example shows.

Example 13.4.6

The set of partial sums of the series 1 − 1 + 1 − 1 + ⋯ has an upper bound, but the series does not converge.

Proof

In Example 13.1.7, we showed that this series diverges. Since the set of partial sums is simply {0, 1}, 1 is an upper bound for the set of partial sums. □

13.5 The Comparison Test

In the case where all of the terms of an infinite series are nonnegative, the converse of Theorem 13.4.5 is true.

Theorem 13.5.1

If a1 + a2 + a3 + ⋯ is an infinite series with ai ≥ 0 for all i, then the series converges if and only if the set of all partial sums has an upper bound.

Proof

Convergence implies that the set of partial sums has an upper bound, by Theorem 13.4.5. The proof of the converse requires the Least Upper Bound Property (13.4.3). To begin the proof, assume that the set of partial sums has an upper bound. Then, by the Least Upper Bound Property, the set of partial sums has a least upper bound. Call this least upper bound S; we prove that the series converges to S. For this, we must show (Definition 13.1.4) that, for every ε > 0, there is an N such that |Sn −S| < ε whenever n ≥ N.

Let ε > 0 be given. Since ε is greater than 0, S − ε < S. Since S is the least upper bound of the set of partial sums and S − ε is less than S, S − ε cannot be an upper bound of the set of partial sums. Thus, there is some N such that S − ε < SN. We show that this N satisfies the definition of convergence (13.1.4) for the given ε. The hypothesis that each term of the series is nonnegative implies that if n1 ≤ n2, then $$\textit {S}_{n_1}\leq \textit {S}_{n_2}$$, since $$\textit {S}_{n_2}$$ is obtained by adding nonnegative numbers to $$\textit {S}_{n_1}$$. In particular, SN ≤Sn whenever n ≥ N. Also, since S is an upper bound for the set of partial sums, Sn ≤S for every n. Therefore, when n ≥ N,
\displaystyle \begin{aligned}\textit{S}- \varepsilon < \textit{S}_N \leq \textit{S}_n \leq \textit{S}\end{aligned}
In particular, S − ε < Sn < S + ε for every such n. Subtracting S from each term in this inequality yields − ε < Sn −S < ε, which is equivalent to |Sn −S| < ε. Thus, Sn is within ε of S for every n ≥ N, and the theorem follows. □

Example 13.5.2

The series
\displaystyle \begin{aligned}\frac{1}{1 \cdot 2} + \frac{1}{2\cdot 2^2} + \frac{1}{3 \cdot 2^3} + \frac{1}{4 \cdot 2^4} + \cdots + \frac{1}{n \cdot 2^n} + \cdots\end{aligned}
converges.

Proof

First, $$\frac {1}{n\cdot 2^n}$$ is less than or equal to $$\frac {1}{2^n}$$ for every natural number n. Thus, every partial sum of this series is less than or equal to the corresponding partial sum of the geometric series $$\frac {1}{2} + \frac {1}{2^2} + \frac {1}{2^3} + \cdots$$. Since the sum of the geometric series with ratio $$\frac {1}{2}$$ and first term $$\frac {1}{2}$$ is 1 (Example 13.2.5), it follows that the set of partial sums of both series have 1 as an upper bound. Therefore, by Theorem 13.5.1, the series converges. □

The previous example is a special case of a general situation: A series with nonnegative terms must converge if it is “term by term” less than or equal to a convergent series. That is, the following theorem holds.

The Comparison Test 13.5.3

Suppose 0 ≤ an ≤ bn for every natural number n.
1. (i)

If the series b1 + b2 + b3 + ⋯ converges, then the series a1 + a2 + a3 + ⋯ converges.

2. (ii)

If the series a1 + a2 + a3 + ⋯ diverges, then the series b1 + b2 + b3 + ⋯ diverges.

Proof

(i) Let S be the sum of the series b1 + b2 + b3 + ⋯. It is clear that every partial sum of the series a1 + a2 + a3 + ⋯ is at most S. Thus, by Theorem 13.5.1, the series a1 + a2 + a3 + ⋯ converges.

(ii) If b1 + b2 + b3 + ⋯ did converge, then a1 + a2 + a3 + ⋯ would converge, by part (i). Therefore, b1 + b2 + b3 + ⋯ diverges. □

The following is a very easy application of the Comparison Test.

Example 13.5.4

The series $$\frac {1}{1^2 \cdot 7} + \frac {1}{2^2 \cdot 7^2} + \frac {1}{3^2 \cdot 7^3} + \frac {1}{4^2 \cdot 7^4} + \cdots + \frac {1}{n^2 \cdot 7^n} + \cdots$$ converges.

Proof

This series is clearly term by term less than the convergent geometric series $$\frac {1}{7} + \frac {1}{7^2} + \frac {1}{7^3} + \cdots$$. □

A slightly more complicated application is the following.

Example 13.5.5

The series
\displaystyle \begin{aligned}\frac{1}{3} + \frac{2}{3^2} + \frac{3}{3^3} + \frac{4}{3^4} + \cdots + \frac{n}{3^n} + \cdots \end{aligned}
converges.

Proof

We will establish that $$\frac {n}{3^n}$$ is less than $$\frac {1}{2^n}$$ for every natural number n; the convergence of the given series then follows from part (i) of the Comparison Test (13.5.3) by comparing the given series to the geometric series $$\frac {1}{2} + \frac {1}{4} + \frac {1}{8} + \cdots$$.

The fact that $$\frac {n}{3^n}$$ is less than $$\frac {1}{2^n}$$ for every natural number n can be established by the Generalized Principle of Mathematical Induction (). To see this, note that $$\frac {n}{3^n} < \frac {1}{2^n}$$ is equivalent to $$\left (\frac {2}{3}\right )^n < \frac {1}{n}$$. This is true for n = 1 and also for n = 2. Now suppose that $$\left (\frac {2}{3}\right )^k < \frac {1}{k}$$ for some k ≥ 2. Note that $$\frac {2}{3} \leq \frac {k}{k+1}$$ when k ≥ 2. It follows that
\displaystyle \begin{aligned}\left(\frac{2}{3}\right)^{k+1} = \left(\frac{2}{3}\right)^{k} \cdot \frac{2}{3} < \frac{1}{k} \cdot \frac{k}{k+1} = \frac{1}{k+1}\end{aligned}

Therefore, $$\frac {n}{3^n} < \frac {1}{2^n}$$ for all n ≥ 2. Thus, part (i) of the Comparison Test (13.5.3) gives the result. □

13.6 Representing Real Numbers by Infinite Decimals

We show that every infinite decimal represents a real number, and, conversely, that every real number has a representation as an infinite decimal.

We define nonnegative infinite decimals as follows.

Definition 13.6.1

A nonnegative infinite decimal is an expression of the form M.a1a2a3…, where M is a nonnegative integer and each ai is a “digit” (i.e., a number in the set {0, 1, …, 9}). We interpret such an expression as representing the infinite series $$M + \frac {a_1}{10} + \frac {a_2}{10^2} + \cdots +\frac {a_k}{10^k} + \cdots$$.

Theorem 13.6.2

Every nonnegative infinite decimal converges.

Proof

We must show that the infinite series $$M + \frac {a_1}{10} + \frac {a_2}{10^2} + \cdots +\frac {a_k}{10^k} + \cdots$$ converges whenever M is a nonnegative integer and each ak is a digit. Since each ak is a digit, $$\frac {a_k}{10^k}$$ is less than or equal to $$\frac {9}{10^k}$$, for every k. It follows that the infinite decimal is term by term less than or equal to the “comparison series”
\displaystyle \begin{aligned}M + \frac{9}{10} + \frac{9}{10^2} + \frac{9}{10^3} + \cdots + \frac{9}{10^k} + \cdots\end{aligned}
This comparison series converges to $$M + \frac {\left (\frac {9}{10}\right )} {\left (1 - \frac {1}{10}\right )}= M + 1$$, since its partial sums are all of the form M plus a partial sum of the convergent geometric series $$\frac {9}{10} + \frac {9}{10^2} + \frac {9}{10^3} + \cdots$$. Therefore, part (i) of the Comparison Test (13.5.3) gives the result. □

Negative infinite decimals can be defined as negations of nonnegative ones. For example, − 7.11… = −(7.11… ). In general, when M is a nonnegative integer and each ak is a digit, we define − M.a1a2a3… to be $$(-1)\cdot \big (M + \frac {a_1}{10} + \frac {a_2}{10^2} + \frac {a_3}{10^3} + \cdots \big )$$.

Theorem 13.6.3

Every real number can be represented as an infinite decimal .

Proof

First, the infinite decimal .0000… represents the real number 0. We next consider the case where r is a positive real number. We can obtain a decimal representation for r as follows. Let M be the greatest integer that is less than or equal to r. Such an M will be 0 or a natural number. Note that r − M is less than 1, since otherwise M + 1 would be an integer less than or equal to r. Let a1 be the largest digit such that $$\frac {a_1}{10}$$ is less than or equal to r − M. Then $$r - \big (M + \frac {a_1}{10}\big )$$ is less than $$\frac {1}{10}$$. Let a2 be the largest digit such that $$\frac {a_2}{10^2}$$ is less than or equal to $$r - \big (M + \frac {a_1}{10}\big )$$. Then $$r - \big (M + \frac {a_1}{10} + \frac {a_2}{10^2}\big )$$ is less than $$\frac {1}{10^2}$$; then let a3 be the largest digit such that $$\frac {a_3}{10^3}$$ is less than $$r - \big (M + \frac {a_1}{10} + \frac {a_2}{10^2}\big )$$.

Continue constructing digits ak in this manner. Then, for each k, the absolute value of $$r -\big (M + \frac {a_1}{10} + \frac {a_2}{10^2} + \cdots + \frac {a_k}{10^k}\big )$$ is less than $$\frac {1}{10^k}$$. We claim that the infinite series $$M + \frac {a_1}{10} + \frac {a_2}{10^2} + \frac {a_3}{10^3} + \cdots$$ converges to r. To see this, let ε > 0 be given and choose N large enough that 10N−1 is greater than $$\frac {1}{\varepsilon }$$. If Sn is the nth partial sum of the series $$M + \frac {a_1}{10} + \frac {a_2}{10^2} + \frac {a_3}{10^3} + \cdots$$ and n ≥ N, then
\displaystyle \begin{aligned} |r-\textit{S}_n|=\left|\, r - \left(M + \frac{a_1}{10} + \frac{a_2}{10^2} + \cdots + \frac{a_{n-1}}{10^{n-1}}\right)\,\right|< \frac{1}{10^{n-1}} \leq \frac{1}{10^{N-1}}< \varepsilon \end{aligned}
Thus, the series converges to r.

If r is a negative real number, an infinite decimal representation of r can be obtained by multiplying an infinite decimal representation of − r by − 1. □

Thus, real numbers can be represented by infinite decimals. However, the representation of a real number by an infinite decimal is not necessarily unique. Some real numbers have two distinct representations. For example, summing the corresponding infinite series shows that .299999… equals .300000…. More generally, every real number represented by an infinite decimal that ends with an infinite string of 9’s also has a representation that ends with a string of 0’s. That is the only way that a real number can have two distinct infinite decimal representations. Before proving this fact, we make the following observation.

Lemma 13.6.4

The infinite decimal .c1c2c3is at most 1. Moreover, if any ck is less than 9, then .c1c2c3is less than 1. In particular, .c1c2c3is equal to 1 if and only if ck = 9 for every k.

Proof

The infinite series that .c1c2c3… represents is term by term less than or equal to the geometric series $$\frac {9}{10} + \frac {9}{10^2} + \frac {9}{10^3} + \cdots$$. Since that geometric series sums to 1, every partial sum of $$\frac {c_1}{10} + \frac {c_2}{10^2} + \frac {c_3}{10^3} + \cdots$$ is at most 1. Therefore, the infinite decimal .c1c2c3… is at most 1. If ck is strictly less than 9, then every partial sum of $$\frac {c_1}{10} + \frac {c_2}{10^2} + \frac {c_3}{10^3} + \cdots$$ is less than $$1 - \frac {1}{10^k}$$, so .c1c2c3… is at most $$1 - \frac {1}{10^k}$$. □

Theorem 13.6.5

If two different infinite decimals represent the same real number, then one of them ends in a string of 9’s and the other ends in a string of 0’s.

Proof

Clearly, it suffices to prove the theorem for representations of positive real numbers. Suppose, then, that a positive number has distinct representations
\displaystyle \begin{aligned}M.a_1a_2a_3\ldots = N.b_1b_2b_3\dots\end{aligned}
(Saying that the representations are distinct means that they differ in at least one digit.) Consider the case where M and N are different. Since M ≠ N, one of them is larger. Suppose that M is greater than N; that is, M − N ≥ 1. On the other hand, M − N ≤ 1 since
\displaystyle \begin{aligned} M-N\leq (M-N)+.a_1a_2a_3\ldots = M.a_1a_2a_3\ldots - N = .b_1b_2b_3\ldots \end{aligned}
which we know from Lemma 13.6.4 is at most 1. Thus, M − N = 1, and the above gives
\displaystyle \begin{aligned} 1\leq 1+.a_1a_2a_3\ldots =.b_1b_2b_3\ldots \leq 1 \end{aligned}
This implies that .b1b2b3… = 1 and .a1a2a3⋯ = 0. Therefore, every ak is 0 and, by Lemma 13.6.4, every bk is 9.
Now suppose that M = N and that the nth decimal place is the first decimal place where the two representations differ; that is, aj = bj for all j less than n and an is different from bn. Multiplying by 10n yields distinct representations
\displaystyle \begin{aligned} a_n.a_{n+1}a_{n+2}a_{n+3}\ldots = b_n.b_{n+1}b_{n+2}b_{n+3}\ldots \end{aligned}
Since an ≠ bn, one of them is larger; suppose that an is greater than bn. Then, the first case implies that ak = 0 and bk = 9 for every k ≥ n + 1. This proves the theorem. □

One possible way of constructing the real numbers from the integers is to use infinite decimals. However, it is not easy to describe the basic arithmetic operations of addition and multiplication in terms of infinite decimals. Another way of constructing the real numbers is outlined in Problem in Chapter .

13.7 Further Examples of Infinite Series

Definition 13.7.1

The harmonic series is the series
\displaystyle \begin{aligned}1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots + \frac{1}{n} + \cdots\end{aligned}

Theorem 13.7.2

The harmonic series diverges.

Proof

This will follow from Theorem 13.4.5 if we show that the set of partial sums does not have an upper bound. We do this by establishing that, for every natural number M, there is a partial sum that is greater than $$\frac {1}{2}\cdot M$$.

Begin by observing that $$\frac {1}{3} + \frac {1}{4}$$ is greater than $$\frac {1}{4} + \frac {1}{4}=\frac {1}{2}$$. Similarly, $$\frac {1}{5} + \frac {1}{6} + \frac {1}{7} + \frac {1}{8}$$ is greater than $$\frac {1}{8} + \frac {1}{8} + \frac {1}{8} + \frac {1}{8} = \frac {1}{2}$$. Moreover,
\displaystyle \begin{aligned}\frac{1}{9} + \frac{1}{10} + \cdots + \frac{1}{16}\end{aligned}
is greater than
\displaystyle \begin{aligned}\frac{1}{16} + \frac{1}{16} + \cdots + \frac{1}{16} = \frac{8}{16} = \frac{1}{2}\end{aligned}
In general, for every natural number k, each of the terms of the harmonic series from $$\frac {1}{2^{k-1} + 1}$$ to $$\frac {1}{2^k}$$ is at least $$\frac {1}{2^{k}}$$, and there are 2k − 2k−1 = 2k−1 ⋅ (2 − 1) = 2k−1 such terms. Therefore, the contribution of the sum of those terms to the harmonic series is at least $$2^{k-1}\cdot \frac {1}{2^{k}}=\frac {1}{2}$$.

The above shows that, for every natural number k, the partial sum $$\textit {S}_{2^{k}}$$ is at least $$\frac {1}{2}\cdot k$$. Thus, for every natural number M, there are partial sums of the harmonic series which are greater than $$\frac {1}{2}\cdot M$$, so the series diverges. □

Example 13.7.3

The series
\displaystyle \begin{aligned}1 + \frac{1}{\sqrt {2}} + \frac{1}{\sqrt {3}} + \cdots + \frac{1}{\sqrt {n}} + \cdots\end{aligned}
diverges.

Proof

For every natural number n, $$\sqrt {n} \leq n$$, so $$\frac {1}{\sqrt {n}} \geq \frac {1}{n}$$. Therefore, the given series diverges by part (ii) of the Comparison Test (13.5.3), comparing it to the harmonic series (Theorem 13.7.2). □

Example 13.7.4

The series
\displaystyle \begin{aligned}\frac{1}{1\cdot 2}+ \frac{1}{2 \cdot 3} + \cdots + \frac{1}{n(n+1)} + \cdots\end{aligned}
converges to 1.

Proof

By Problem in Chapter , the nth partial sum of this series is $$\frac {n}{n+1}$$. It is apparent that this is close to 1 if n is large. To formally establish this, let ε be any positive number. If N is a natural number such that $$N +1 > \frac {1}{\varepsilon }$$, then $$\varepsilon > \frac {1}{N+1}$$. Therefore, if n ≥ N, then $$\big |1 - \frac {n}{n+1}\big | = \frac {1}{n+1} \leq \frac {1}{N+1} < \varepsilon$$. □

Example 13.7.5

The infinite series
\displaystyle \begin{aligned}1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots + \frac{1}{n^2} + \cdots\end{aligned}
converges.

Proof

Since the partial sums of
\displaystyle \begin{aligned}1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots + \frac{1}{n^2} + \cdots\end{aligned}
are obtained by adding 1 to those of
\displaystyle \begin{aligned}\frac{1}{2^2} + \frac{1}{3^2} + \cdots + \frac{1}{n^2} + \cdots\end{aligned}
it suffices to show that this latter series converges. We will establish this by comparison to the convergent series from Example 13.7.4. The series
\displaystyle \begin{aligned}\frac{1}{2^2} + \frac{1}{3^2} + \cdots + \frac{1}{(n+1)^2} + \cdots\end{aligned}
is term by term less than the series
\displaystyle \begin{aligned}\frac{1}{1\cdot 2}+ \frac{1}{2 \cdot 3} + \cdots + \frac{1}{n(n+1)} + \cdots\end{aligned}
since $$\frac {1}{(n+1)^2}$$ is less than $$\frac {1}{n(n+1)}$$ for every natural number n. Therefore, the series
\displaystyle \begin{aligned}\frac{1}{2^2} + \frac{1}{3^2} + \cdots + \frac{1}{(n+1)^2} + \cdots\end{aligned}
converges by the Comparison Test (13.5.3). □

It can be shown, using calculus, that the series in the above example converges to $$\frac {\pi ^2}{6}$$.

Theorem 13.7.6

For every p ≥ 2, the series
\displaystyle \begin{aligned}1 + \frac{1}{2^p} + \frac{1}{3^p} + \cdots + \frac{1}{n^p} + \cdots\end{aligned}

converges.

Proof

This follows immediately from the previous example and the fact that p ≥ 2 implies that $$\frac {1}{n^p} \leq \frac {1}{n^2}$$ for every natural number n. □

It can be proven (using integral calculus) that, in fact, $$1 + \frac {1}{2^p} + \frac {1}{3^p} + \cdots$$ converges for every p > 1.

Infinite series are often used to define specific numbers. For example, the famous number e, the base of the natural logarithm, can be defined as an infinite series. (It can also be defined in many other ways.)

Example 13.7.7

The series $$1 + \frac {1}{1!} + \frac {1}{2!} + \frac {1}{3!} + \cdots + \frac {1}{n!} + \cdots$$ converges. The sum of the series is denoted by e .

Proof

An easy application of the Generalized Principle of Mathematical Induction shows that n! > 2n for all n ≥ 4 (see Theorem of Chapter ). Thus, $$\frac {1}{n!} < \frac {1}{2^n}$$ for all n ≥ 4. Hence, $$\frac {1}{4!} + \frac {1}{5!} + \frac {1}{6!} + \cdots$$ converges by the Comparison Test (13.5.3), and therefore so does the entire series. □

We next consider a particularly interesting example of a divergent series.

Theorem 13.7.8

Let pj denote the jth prime number (so that p1 = 2, p2 = 3, p3 = 5, etc.). Then the series
\displaystyle \begin{aligned}\frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{p_j} + \cdots\end{aligned}

diverges. That is, the series of the reciprocals of the primes diverges.

Proof

This can be proven in several ways. The proof that we present below appears to be the easiest, although it is somewhat tricky. We show that the assumption that the series converges leads to a contradiction.

Suppose that the sum of the series was S. Then, by the definition of convergence (Definition 13.1.4), there would be a partial sum of the series, say Sk, such that Sk is within $$\frac {1}{2}$$ of S. It would then follow that the sum of the series obtained by discarding the first k terms would be less than $$\frac {1}{2}$$. That is, there would be a natural number k such that
\displaystyle \begin{aligned}\frac{1}{p_{k+1}} + \frac{1}{p_{k+2}} + \frac{1}{p_{k+3}} + \cdots\end{aligned}
is less than $$\frac {1}{2}$$. We proceed to show that this is impossible.

For the rest of the proof, we fix such a k and say that pj is a “small prime” if j is less than or equal to k and that pj is a “big prime” if j is greater than k. For each natural number x, let N(x) denote the number of natural numbers that are less than or equal to x and are not divisible by any big prime. The surprising trick involves obtaining, and using, an upper bound on N(x).

Fix any natural number x. We can get a crude upper bound for N(x) as follows. Every natural number y that is counted in N(x) can be written as a product uv, where u is a perfect square and v does not have any perfect square divisors. To see this, use the prime factorization (Corollary ) of y to factor out the biggest perfect square u; v is the other factor of y. Note that u = 1 if 1 is the largest perfect square that divides y, and v = 1 if y itself is a perfect square. The number of distinct u’s that arise from y’s that are counted in N(x) must be less than or equal to $$\sqrt {x}$$, since each u is less than or equal to x and is therefore the square of a number that is less than or equal to $$\sqrt {x}$$. Also, each of the v’s consists of a product of small primes raised to at most the first power. There are k small primes, so the number of possible v’s is at most 2k (since each small prime may or may not occur in the prime factorization of each v). Every y that is counted by N(x) is of the form uv and there are at most $$\sqrt {x}$$ u’s and 2k v’s, from which it follows that $$N(x) \leq 2^k \sqrt {x}$$, for each x. We will use this inequality to derive a contradiction.

Next, we establish a lower bound for N(x) that will be inconsistent with the above inequality when x is large. First, for any prime p and any natural number x, there are at most $$\frac {x}{p}$$ multiples of p that are less than or equal to x. Thus there are at most $$\frac {x}{p_j}$$ natural numbers less than or equal to x that have the big prime pj as a factor. Remember that N(x) denotes the number of natural numbers less than or equal to x that have only small prime factors. Therefore, x − N(x) is the number of natural numbers less than or equal to x having at least one big prime as a factor. There are at most $$\frac {x}{p_{k+1}}$$ of those that have the big prime pk+1 as a factor; there are at most $$\frac {x}{p_{k+2}}$$ of those that have the big prime pk+2 as a factor; and so on. Thus, x − N(x) is less than or equal to $$\frac {x}{p_{k+1}} + \frac {x}{p_{k+2}} + \frac {x}{p_{k+3}} + \cdots$$. Each partial sum of this series is x times a partial sum of the series $$\frac {1}{p_{k+1}} + \frac {1}{p_{k+2}} + \frac {1}{p_{k+3}} + \cdots$$. Since the sum of the latter series is less than $$\frac {1}{2}$$, it follows that x − N(x) is less than $$\frac {x}{2}$$.

The inequality $$x - N(x) < \frac {x}{2}$$ is equivalent to $$\frac {x}{2} < N(x)$$. Combining this with the previous inequality we obtained for N(x) gives
\displaystyle \begin{aligned}\frac{x}{2} < N(x) \leq 2^k\sqrt {x} \end{aligned}
Therefore,
\displaystyle \begin{aligned}\frac{x}{2} < 2^k \sqrt{x}\end{aligned}
Multiplying both sides of this inequality by 2 and dividing by $$\sqrt {x}$$ yields
\displaystyle \begin{aligned}\sqrt {x} < 2^{k+1}\end{aligned}

Thus, assuming that the series converges leads to the conclusion that there is a natural number k such that $$\sqrt {x} < 2^{k+1}$$ for every natural number x. Now we have our contradiction: Since k is fixed, the above cannot hold for all natural numbers x. For example, if x = 22k+4, then $$\sqrt {x} = 2^{k+2}$$, which is larger than 2k+1. □

The preceding provides another proof that there are infinitely many primes; if there were only a finite number of primes, the series consisting of the reciprocals of the primes would have only a finite number of terms, and therefore would converge.

In Chapter , we mentioned the famous unsolved twin primes problem. While (in spite of dramatic progress by Yitang Zhang and others, beginning in 2013) the twin primes problem is still unsolved, Viggo Brun proved in 1919 that the sum of the reciprocals of the twin primes converges. (The proof of this is well beyond the scope of this book.)

There is much more that is known about infinite series. Some other results are outlined in the interesting and challenging problems below. Almost every calculus book contains a large amount of related material.

13.8 Problems

Basic Exercises

1. 1.
Find the sum of each of the following geometric series:
1. (a)

$$1 - \frac {1}{3} + \frac {1}{3^2} - \frac {1}{3^3} + \cdots$$

2. (b)

$$10 + \frac {10}{7} + \frac {10}{7^2} + \frac {10}{7^3} + \cdots$$

3. (c)

$$\frac {3}{\sqrt {2}} + \frac {3}{\big (\sqrt {2}\big )^2} + \frac {3}{\big (\sqrt {2}\big )^3} + \frac {3}{\big (\sqrt {2}\big )^4} + \cdots$$

2. 2.
For each of the following sets, determine if the set has an upper bound, and, if so, find the least upper bound:
1. (a)

{1, −1, 2, −2, 3, −3, … }

2. (b)

$$\{1, -1, \frac {1}{2}, -\frac {1}{2}, \frac {1}{3}, -\frac {1}{3}, \dots \}$$

3. (c)

$$\{-1, - \frac {1}{2}, - \frac {1}{3}, -\frac {1}{4}, \dots \}$$

4. (d)

{−7, 12, −4, 6}

3. 3.
Determine which of the following series converges:
1. (a)

$$1 - \frac {1}{7} + \frac {1}{7^2} - \frac {1}{7^3} + \frac {1}{7^4} - \cdots$$

2. (b)

$$\sqrt {2} + 789 + 2042 + \frac {3}{2} + \frac {3}{2^2} + \frac {3}{2^3} + \cdots + \frac {3}{2^n} + \cdots$$

3. (c)

.1 + .1 + .1 + .1 + ⋯

4. (d)

$$1 + \frac {1}{2} + \frac {1}{3} + \frac {1}{2} + \frac {1}{4} + \frac {1}{2} + \frac {1}{5} + \frac {1}{2} + \cdots$$

5. (e)

$$1 + \frac {1}{2} + \frac {1}{2^2} + \frac {1}{2} + \frac {1}{2^3} + \frac {1}{2} + \frac {1}{2^4} + \frac {1}{2} + \cdots$$

6. (f)

$$\frac {5}{3^2} + \frac {17}{3^3} + \frac {5}{3^4} + \frac {17} {3^5} + \cdots$$

4. 4.

For each of the following infinite decimals, determine the rational number that it represents:

1. (a)

.77777777…

2. (b)

.3434343434…

3. (c)

17.389389389389…

5. 5.

Let t be any positive real number. Use the Principle of Mathematical Induction to prove that (1 + t)n ≥ 1 + nt for every natural number n. (This result was used without proof in Theorem 13.2.3.)

6. 6.
Determine which of the following series converge:
1. (a)

$$\frac {1}{2} + \frac {2}{3} + \frac {3}{4} + \frac {4}{5} + \frac {5}{6} + \cdots + \frac {n}{n+1} + \cdots$$

2. (b)

$$\frac {1}{5} + \frac {1}{9} + \frac {1}{17} + \cdots + \frac {1}{2^n + 1} + \cdots$$

Interesting Problems

1. 7.

Determine which of the following series converge:

1. (a)

$$19 + \frac {19}{2^{7/2}} + \frac {19}{3^{7/2}} + \frac {19}{4^{7/2}} + \cdots + \frac {19}{n^{7/2}} + \cdots$$

2. (b)

$$\frac {1}{2} + \frac {1}{3} + \frac {1}{2^2} + \frac {1}{3^2} + \frac {1}{2^3} + \frac {1}{3^3} + \cdots + \frac {1}{2^n} + \frac {1}{3^n} + \cdots$$

2. 8.

Determine the rational number that is represented by each of the following infinite decimals:

1. (a)

6.798345345345345345…

2. (b)

− 38.0006561234123412341234…

3. (c)

.012345678901234567890123456789…

3. 9.
Suppose that the sum of the series a1 + a2 + a3 + ⋯ + an + ⋯ is S and the sum of the series b1 + b2 + b3 + ⋯ + bn + ⋯ is T. Prove that the sum of the series
\displaystyle \begin{aligned}(a_1 - b_1) + (a_2 - b_2) + (a_3 - b_3) + \cdots + (a_n - b_n) + \cdots\end{aligned}
is S − T.

4. 10.
(“Sigma notation”) There is a standard notation that is often used when considering infinite series and in many other contexts. The Greek letter $$\sum$$ (called “sigma”) is part of a shorthand for representing sums, as illustrated by the following examples:
1. (i)

$$\sum \limits _{i = 1}^4 a_i$$ is defined to be a1 + a2 + a3 + a4; it can be read “the sum from i = 1 to 4 of ai

2. (ii)

$$\sum \limits _{n=3}^5 \frac {n}{2}$$ means $$\frac {3}{2} + \frac {4}{2} + \frac {5}{2}$$; it can be read “the sum from n = 3 to 5 of $$\frac {n}{2}$$

3. (iii)

$$\sum \limits _{j = 5}^{21} j^2$$ means 52 + 62 + 72 + ⋯ + 212; it can be read “the sum from j = 5 to 21 of j2

4. (iv)

$$\sum \limits _{i = 1}^{\infty } \frac {1}{2^i}$$ means $$\frac {1}{2} + \frac {1}{2^2} + \frac {1}{2^3} + \cdots + \frac {1}{2^i} + \cdots$$; it can be read “the sum from i = 1 to infinity of $$\frac {1}{2^i}$$

Thus, $$\sum$$ is used as above to indicate sums. When we write $$\sum \limits _{i = 1}^{\infty } a_i$$ we do not necessarily imply that the series converges; it is merely shorthand for the infinite series a1 + a2 + a3 + ⋯. If the series converges to S, we may write $$\sum \limits _{i=1}^{\infty } a_i = S$$. For example, $$\sum \limits _{i = 1}^{\infty } \frac {1}{2^i} = 1$$.
1. (a)

Find: $$\sum \limits _{i=1}^4 \frac {1}{i}$$

2. (b)

Find: $$\sum \limits _{i=3}^7 (-1)^i \cdot (i + 4)$$

3. (c)

Find: $$\sum \limits _{i=1}^\infty \frac {7}{4^i}$$

4. (d)

Find: $$\sum \limits _{i=5}^\infty (-1)^i \cdot \frac {19}{5^i}$$

5. (e)

Find: $$\sum \limits _{i=1}^{100} (-1)^i$$

6. (f)

Find: $$\sum \limits _{i=2}^{17} \left (\frac {1}{i} - \frac {1}{i + 1}\right )$$

7. (g)

Show that $$m\sum \limits _{i = 1}^n a_i = \sum \limits _{i=1}^n ma_i$$, where n is any natural number and m is any real number.

8. (h)

Show that $$\left (\sum \limits _{i = 1}^n a_i\right ) + \left (\sum \limits _{i = 1}^n b_i\right ) = \sum \limits _{i = 1}^n (a_i + b_i)$$.

5. 11.

Show that there is no least upper bound property for the set of rational numbers. In other words, show that there are nonempty sets of rational numbers that have rational upper bounds but which do not have a least rational upper bound.

Challenging Problems

1. 12.

Show that the series $$1 + \frac {1}{3} + \frac {1}{5} + \frac {1}{7} + \cdots$$ diverges.

2. 13.
1. (a)

Suppose that the series a1 + a2 + a3 + ⋯ converges. Prove that, for each positive number δ, there is a natural number N such that |ai|≤ δ for every i ≥ N.

2. (b)

Suppose that there is a positive number δ such that |ai|≥ δ for infinitely many ai. Show that the series a1 + a2 + a3 + ⋯ diverges.

3. (c)

Let $$a_i = (-1)^i \frac {i}{i + 100}$$. Show that the series a1 + a2 + a3 + ⋯ diverges.

3. 14.
(A form of the “Ratio Test”)
1. (a)

Show that the series a1 + a2 + a3 + ⋯ with nonnegative terms converges if there is a positive number r < 1 such that ai+1 is less than or equal to rai for every i.

• [Hint: Compare the given series to the geometric series a1 + a1r + a1r2 + a1r3 + ⋯.]

1. (b)

Show that the series a1 + a2 + a3 + ⋯ with nonnegative terms converges if there exist an N and a positive number r < 1 such that ai+1 is less than or equal to rai for every i ≥ N.

4. 15.

(Absolute convergence implies convergence) The series a1 + a2 + a3 + ⋯ is said to converge absolutely if the series |a1| + |a2| + |a3| + ⋯ converges. The following is an outline of a proof that a series converges if it converges absolutely.

Suppose that |a1| + |a2| + |a3| + ⋯ converges.

1. (a)

Show that 2|a1| + 2|a2| + 2|a3| + ⋯ converges.

2. (b)

Use the Comparison Test (13.5.3) to show that (|a1| + a1) + (|a2| + a2) + (|a3| + a3) + ⋯ converges.

3. (c)

Prove that the series −|a1|−|a2|−|a3|−⋯ converges, and add it to the series in part (b) to show that a1 + a2 + a3 + ⋯ converges.

5. 16.

Prove that, for every real number x, the series $$1 + x + \frac {x^2}{2!} + \frac {x^3}{3!} + \frac {x^4}{4!} + \cdots$$ converges. (This is one definition of the exponential function ex, where e is the base of the natural logarithm. That is, $$e^x = 1 + x + \frac {x^2}{2!} + \frac {x^3}{3!} + \frac {x^4}{4!} + \cdots$$.)

• [Hint: Since absolute convergence implies convergence (Problem 15), it suffices to prove convergence for positive x. For this, use the Ratio Test (Problem 14) with N any natural number larger than x and with $$r = \frac {x}{x + 1}$$.]

1. 17.

(A form of the “Root Test”) Suppose that a1 + a2 + a3 + ⋯ is a series with nonnegative terms. Prove that the series converges if there is a positive number r < 1 and a natural number N such that $$(a_i)^{\frac {1}{i}} \leq r$$ for all i ≥ N.

2. 18.
The alternating harmonic series is the series:
\displaystyle \begin{aligned}1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{(-1)^{n+1}}{n} + \cdots\end{aligned}
1. (a)

Let $$\mathcal {T}$$ denote the set of even partial sums of the alternating harmonic series; that is, $$\mathcal {T} = \{1 - \frac {1}{2}, 1 - \frac {1}{2} + \frac {1}{3} - \frac {1}{4}, 1 - \frac {1}{2} + \frac {1}{3} - \frac {1}{4} + \frac {1}{5} - \frac {1}{6}, \ldots \}$$. Show that 1 is an upper bound for $$\mathcal {T}$$.

2. (b)

Let S be the least upper bound of $$\mathcal {T}$$. Show that the alternating harmonic series converges to S.

3. 19.
1. (a)

Show that the terms of the alternating harmonic series $$1 - \frac {1}{2} + \frac {1}{3} - \frac {1}{4} + \cdots$$ can be rearranged so that the resulting series converges to 7.

• [Hint: Use the divergence of the series $$1 + \frac {1}{3} + \frac {1}{5} + \cdots$$ (Problem 12) to get the first partial sum that is greater than 7. Begin the rearrangement with that partial sum. Then, adding the first negative term, $$-\frac {1}{2}$$, will make the sum less than 7. Then add positive terms until the result is just greater than 7; then add negative terms to get less than 7, and so on.]

1. (b)

Let t be any real number. Show that the terms of the alternating harmonic series can be rearranged so that the resulting series converges to t.

4. 20.

Give an example of a series which converges but does not converge absolutely (see Problem 15).

5. 21.

A series is said to converge conditionally if it converges but does not converge absolutely. Prove that any conditionally convergent series can be rearranged to sum to any real number, and can also be rearranged so that it diverges.

• [Hint: First show that the sum of the nonnegative terms of the series diverges, as does the sum of the negative terms of the series.]

1. 22.

Prove that, if a series converges absolutely, then all the rearrangements of the series have the same sum.

2. 23.
(Characterization of the rational numbers) A repeating infinite decimal is an infinite decimal of the form:
\displaystyle \begin{aligned}L.a_1a_2\cdots a_{m}b_1b_2\cdots b_nb_1b_2\cdots b_nb_1b_2 \cdots b_n\cdots\end{aligned}
where L is an integer and the ai and bi are digits.
1. (a)

Show that every repeating infinite decimal represents a rational number.

2. (b)

Show that every rational number has a representation as a repeating infinite decimal.

3. 24.

A number t is said to be a lower bound for a set $$\mathcal {S}$$ of real numbers if t is less than or equal to x for every x in $$\mathcal {S}$$. A number t0 is a greatest lower bound for the set $$\mathcal {S}$$ if t0 is a lower bound for $$\mathcal {S}$$ and t0 is greater than or equal to t for every lower bound t of $$\mathcal {S}$$. Prove that every nonempty set of real numbers that has a lower bound has a greatest lower bound.

• [Hint: Consider the set $$\mathcal {T} = \{-x : x \in \mathcal {S}\}$$.]

1. 25.
1. (a)

Prove that, if a series converges, then the set of all its partial sums has a lower bound (see Problem 13.8).

2. (b)

Prove that a series whose terms are all less than or equal to 0 converges if and only if the set of all its partial sums has a lower bound.

2. 26.

(Construction of least upper bounds) For this problem we assume familiarity with the construction of the real numbers from sets of rational numbers using Dedekind cuts, as outlined in Problem in Chapter . In that context, the real number A is said to be less than or equal to the real number B if A is contained in B.

Suppose that a given nonempty set $$\mathcal {S}$$ of real numbers (which is a set of sets of rational numbers) has an upper bound. That is, there is a real number that is greater than or equal to every real number in $$\mathcal {S}$$. Prove that the union of all of the real numbers in $$\mathcal {S}$$ is a real number and is the least upper bound of $$\mathcal {S}$$.

3. 27.

A more standard approach to convergence of infinite series begins with the definition of convergence of any sequence of real numbers. The definition of convergence that we have given (13.1.4) is the particular case when the sequence is the sequence of partial sums of a series. We have delayed the presentation of the more general definition of convergence of sequences because some people find it more confusing to learn the general definition when they are first exposed to this topic. However, the general definition is required in order to obtain many of the standard results on infinite series, and in many other situations. In this exercise, we state the definition of convergence of an arbitrary sequence of real numbers, and use that definition to reformulate and extend some of the results obtained in the chapter and in the problems given above.

Some examples of sequences are:
\displaystyle \begin{aligned}1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots\end{aligned}
\displaystyle \begin{aligned}\sqrt {3}, \sqrt {4}, \sqrt {5}, \dots\end{aligned}
\displaystyle \begin{aligned}1, -1, 1, -1, 1, -1, \dots\end{aligned}
In general, a sequence of real numbers is a listing of real numbers, one for each natural number. More precisely, a sequence of real numbers can be defined as an assignment of a real number to each natural number (that is, as a function from the natural numbers to the real numbers). For example, the sequence $$1 , \frac {1}{2}, \frac {1}{3}, \frac {1}{4}, \dots$$ is the assignment of the number $$\frac {1}{n}$$ to each natural number n, and the sequence 1, −1, 1, −1, 1, −1, … is the assignment of 1 to every odd natural number and − 1 to every even natural number.

Notation such as x1, x2, x3, x4, … is often used to denote a sequence. Sometimes we abbreviate this as (xn). The crucial concept is that of a limit of a sequence. The rough idea is that a real number L is the limit of the sequence (xn) if xn is close to L when n is large. More accurately, no matter how close we wish to get xn to L, it will be that close if n is sufficiently large. The precise definition is the following.

Definition 13.8.1 The sequence (xn) converges to L (or has limit L) if, for every real number ε greater than 0, there is a natural number N such that |xn − L| < ε whenever n ≥ N. We use the notation limnxn = L to denote the fact that L is the limit of the sequence (xn); this is read “the limit as n approaches infinity of the sequence (xn) is L.” Sometimes the notation (xn) → L is used.

Note that applying this definition to a sequence (Sn) of partial sums of an infinite series yields Definition 13.1.4. That is, an infinite series converges to S if and only if the sequence of its partial sums converges to S. In studying an infinite series a1 + a2 + a3 + ⋯, there are several different sequences that naturally arise. The one that we have discussed so far is the sequence (Sn) of partial sums. Another is the sequence (ai) of terms of the series. It is important not to confuse the two. Limits of some other sequences, including some related to the terms of a series, also play a role.

1. (a)

Prove that, if a series converges, then its sequence of terms converges to 0 (see Problem 13). That is, if a1 + a2 + a3 + ⋯ converges, then limiai = 0.

2. (b)
(“The Ratio Test”) Suppose that a1 + a2 + a3 + ⋯ is a series of non-zero terms and suppose that $$\displaystyle \lim _{i \rightarrow \infty } \left |\frac {a_{i+1}}{a_i}\right | = r$$.
1. (i)

Show that the series a1 + a2 + a3 + ⋯ converges absolutely if r < 1.

• [Hint: See Problems 14 and 15.]

1. (ii)

Show that the series a1 + a2 + a3 + ⋯ diverges if r > 1.

2. (iii)

Give an example of a series that diverges and has r = 1.

3. (iv)

Give an example of a series that converges and has r = 1.

3. (c)
(“The Root Test”) Suppose that $$\displaystyle \lim _{i \rightarrow \infty } |a_i|{ }^{\frac {1}{i}} = r$$.
1. (i)

Show that the series a1 + a2 + a3 + ⋯ converges absolutely if r < 1.

• [Hint: See Problems 15 and 13.8.]

1. (ii)

Show that the series a1 + a2 + a3 + ⋯ diverges if r > 1.

4. (d)

(“The Limit Comparison Test”) Let a1 + a2 + a3 + ⋯ and b1 + b2 + b3 + ⋯ be series whose terms are all positive. Suppose that $$\displaystyle \lim _{i \rightarrow \infty } \frac {a_i}{b_i} = r$$ for some r > 0. Show that the series a1 + a2 + a3 + ⋯ converges if and only if the series b1 + b2 + b3 + ⋯ converges.

5. (e)

Determine which of the following series converge.

• [Hint: It may be useful to use some of the results from this chapter, as well as some of the previous parts of this problem.]

1. (i)

$$-2 + \frac {2}{3} -\frac {2}{4} + \frac {2}{5} - \frac {2}{6} + \cdots$$

2. (ii)

$$\frac {1}{\sqrt {11}} + \frac {2}{\sqrt {11^2}} + \frac {3}{\sqrt {11^3}} + \cdots + \frac {n}{\sqrt {11^n}} + \cdots$$

3. (iii)

$$\frac {1^{100}}{3} + \frac {2^{100}}{3^2} + \frac {3^{100}}{3^3} + \frac {4^{100}}{3^4} + \cdots + \frac {n^{100}}{3^n} + \cdots$$

4. (iv)

$$\frac {1}{1 + 1} + \frac {1}{2 + \frac {1}{2}} + \frac {1}{3 + \frac {1}{3}} + \cdots + \frac {1}{n + \frac {1}{n}} + \cdots$$

5. (v)

$$\frac {1}{7^2 - 3\cdot 7} + \frac {1}{8^2 - 3 \cdot 8} + \frac {1}{9^2 - 3 \cdot 9} + \cdots + \frac {1}{n^2 - 3n} + \cdots$$

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Authors and Affiliations

• Daniel Rosenthal
• 1
• David Rosenthal
• 2
• Peter Rosenthal
• 3