Sharp inequalities related to the constant e

Abstract

The aim of this work is to extend the results obtained by Batir and Cancan in (Int. J. Math. Educ. Sci. Technol. 40(8):1101-1109, 2009).

MSC:26A09, 33B10, 26D99.

1 Introduction and preliminary results

Batir and Cancan [[1], Theorem 2.5] presented the following sharp inequalities:

$$exp(1-\frac{n}{2{(n+c)}^2})\le {(1+\frac{1}{n})}^n<exp(1-\frac{n}{2{(n+d)}^2}),n\ge 1,$$

(1)

where $c=\frac{1}{\sqrt{2-ln4}}-1=0.27649\cdots$ and $d=\frac{1}{3}$. The proof is based on the fact that the function

$$\omega (x)=\frac{1}{\sqrt{2(\frac{1}{x}-ln(1+\frac{1}{x}))}}-x$$

is strictly increasing on $(0,\mathrm{\infty })$, while inequalities (1) follow from $\omega (1)\le \omega (n)<\omega (\mathrm{\infty })$. Such an approach of the problem does not offer good results in the left-hand side inequality (1), when n approaches infinity. As we wish to see (1) as an accurate approximation of the form

$${(1+\frac{1}{n})}^n\approx exp(1-\frac{n}{2{(n+\delta (n))}^2}),$$

(2)

we are interested in finding $\delta (n)$ which gives the best such approximation for large values of n. Moreover, numerical computations show us that for large values of n, the expression ${(1+1/n)}^n$ gets closer to the right-hand side of (1). This fact suggests us that the best approximation (2) is obtained when $\delta (n)$ tends to $1/3$, as $n\to \mathrm{\infty }$. For $\delta (n)=1/3$, we deduce

$${(1+\frac{1}{n})}^n\approx exp(1-\frac{n}{2{(n+\frac{1}{3})}^2}),$$

(3)

but a better result is

$${(1+\frac{1}{n})}^n\approx exp(1-\frac{n}{2{(n+\frac{1}{3}-\frac{1}{12n})}^2}).$$

The rigorous argument is the following theorem, which is also an improvement of the Batir and Cancan inequality (1).

Theorem 1 For every real number $x\ge 1$, we have

$$exp(1-\frac{x}{2{(x+\frac{1}{3}-\frac{1}{12x})}^2})<{(1+\frac{1}{x})}^x<exp(1-\frac{x}{2{(x+\frac{1}{3}-\frac{1}{12x}+\frac{23}{540x^2})}^2}).$$

Proof Let

$$f(x)=1-\frac{x}{2{(x+\frac{1}{3}-\frac{1}{12x})}^2}-xln(1+\frac{1}{x})$$

and

$$g(x)=1-\frac{x}{2{(x+\frac{1}{3}-\frac{1}{12x}+\frac{23}{540x^2})}^2}-xln(1+\frac{1}{x}).$$

We have $f(1)=\frac{17}{25}-ln2=-0.013\cdots <0$, $g(1)=\frac{85,351}{121,801}-ln2=0.0075\cdots >0$ and

$$\begin{array}{c}{f}^{″}(x)=-\frac{16x+384x^2+544x^3+15,440x^4+17,664x^5-1}{x{(x+1)}^2{(2x+1)}^4{(6x-1)}^4},\hfill \\ g^{″}(x)=\frac{A(x-1)}{x{(x+1)}^2{(180x^2-45x+540x^3+23)}^4},\hfill \end{array}$$

where

$$\begin{array}{rcl}A(x)& =& 1,031,414,482,080x+3,366,173,125,800x^2\\ +6,243,358,941,720x^3+7,196,770,443,825x^4\\ +5,277,779,208,000x^5+2,403,240,062,400x^6\\ +620,618,112,000x^7+69,480,990,000x^8+137,506,401,616.\end{array}$$

Evidently, f is concave, g is convex on $[1,\mathrm{\infty })$, with $f(\mathrm{\infty })=g(\mathrm{\infty })=0$, so $f<0$ and $g>0$ on $[1,\mathrm{\infty })$. The proof is completed. □

The same remarks we make on Batir and Cancan’s assertion [[1], Theorem 2.6], which is proven to have some computation errors, since the expression ${(1+1/n)}^{n+1}$ can be approximated for large values of n as

$${(1+\frac{1}{n})}^{n+1}\approx exp(1+\frac{n}{2{(n+\frac{1}{6})}^2}),$$

(4)

but a better result is

$${(1+\frac{1}{n})}^{n+1}\approx exp(1+\frac{n}{2{(n+\frac{1}{6}-\frac{1}{24n})}^2}),$$

as we can see from the following.

Theorem 2 For every real number $x\ge 1$, we have

$$exp(1+\frac{x}{2{(x+\frac{1}{6}-\frac{1}{24x}+\frac{43}{2,160x^2})}^2})<{(1+\frac{1}{x})}^{x+1}<exp(1+\frac{x}{2{(x+\frac{1}{6}-\frac{1}{24x})}^2}).$$

Proof Let

$$\begin{array}{c}u(x)=1+\frac{x}{2{(x+\frac{1}{6}-\frac{1}{24x})}^2}-(x+1)ln(1+\frac{1}{x}),\hfill \\ v(x)=1+\frac{x}{2{(x+\frac{1}{6}-\frac{1}{24x}+\frac{43}{2,160x^2})}^2}-(x+1)ln(1+\frac{1}{x}).\hfill \end{array}$$

We have $u(1)=\frac{113}{81}-2ln2=0.00876\cdots >0$, $v(1)=\frac{8,448,529}{6,115,729}-2ln2=-0.00485\cdots <0$ and

$$\begin{array}{c}{u}^{″}(x)=\frac{16x+832x^3+2,624x^4+132,096x^5-1}{x^2(x+1){(4x+24x^2-1)}^4},\hfill \\ v^{″}(x)=-\frac{B(x-1)}{x^2(x+1){(360x^2-90x+2,160x^3+43)}^4},\hfill \end{array}$$

where

$$\begin{array}{rcl}B(x)& =& 63,717,283,707,480x+222,339,093,374,520x^2\\ +443,153,651,482,080x^3+551,897,650,861,200x^4\\ +439,841,844,403,200x^5+219,097,061,222,400x^6\\ +62,377,579,008,000x^7+7,772,423,040,000x^8+7,984,133,521,441.\end{array}$$

Evidently, v is concave, u is convex, with $u(\mathrm{\infty })=v(\mathrm{\infty })=0$, so $v<0$ and $u>0$ on $[1,\mathrm{\infty })$. The proof is completed. □

2 Some extensions

In this section we discuss the problem of approximating ${(1+1/n)}^{n+a}$, $a\in [0,1]$, in the form

$${(1+\frac{1}{n})}^{n+a}\approx exp(1-\frac{n}{p{(n+q)}^2}),p,q\in \mathbb{R}.$$

More precisely, we propose as an open problem the following approximation formula:

$${(1+\frac{1}{n})}^{n+a}\approx exp(1-\frac{(\frac{1}{2}-a)n}{{(n+\frac{3a-2}{6(2a-1)})}^2}),a\in [0,1]\setminus \left\{\frac{1}{2}\right\}.$$

(5)

Particular values $a=0$ and $a=1$ in (5) lead again to (3) and (4).

The special case $a=1/2$ is treated at the final part of this section.

By now, we proved the approximation formula (5) for

$$a\in [0,\frac{4-\sqrt{2}}{7})\cup (\frac{1}{2},\frac{4+\sqrt{2}}{7})\cup (\theta ,1],$$

where $\theta =0.82462\cdots$ is the unique real root of $558a-1,098a^2+675a^3-92$.

This assertion is sustained by the following three theorems.

Theorem 3 Let $a\in [0,\frac{4-\sqrt{2}}{7})$. Then there exists $x_1>0$ (depending on a) such that for every real number $x\ge x_1$, the following inequalities hold true:

$$exp(1-\frac{(\frac{1}{2}-a)x}{{(x+\frac{3a-2}{6(2a-1)}-\frac{7a^2-8a+2}{24{(2a-1)}^{2}x})}^2})<{(1+\frac{1}{x})}^{x+a}<exp(1-\frac{(\frac{1}{2}-a)x}{{(x+\frac{3a-2}{6(2a-1)})}^2}).$$

(6)

Theorem 4 Let $a\in (\frac{1}{2},\frac{4+\sqrt{2}}{7})$. Then there exists $x_2>0$ (depending on a) such that, for every real number $x\ge x_2$, the following inequalities hold true:

$$exp(1+\frac{(a-\frac{1}{2})x}{{(x+\frac{3a-2}{6(2a-1)}+\frac{8a-7a^2-2}{24{(2a-1)}^{2}x})}^2})<{(1+\frac{1}{x})}^{x+a}<exp(1+\frac{(a-\frac{1}{2})x}{{(x+\frac{3a-2}{6(2a-1)})}^2}).$$

(7)

Theorem 5 Let $a\in (\theta ,1]$. Then there exists $x_3>0$ (depending on a) such that, for every real number $x\ge x_3$, the following inequalities hold true:

$$exp(1+\frac{(a-\frac{1}{2})x}{{(x+\frac{3a-2}{6(2a-1)})}^2})<{(1+\frac{1}{x})}^{x+a}<exp(1+\frac{(a-\frac{1}{2})x}{{(x+\frac{3a-2}{6(2a-1)}-\frac{7a^2-8a+2}{24{(2a-1)}^{2}x})}^2}).$$

(8)

Inequalities (6)-(8) are closely related to the functions

$$s(x)=1+\frac{(a-\frac{1}{2})x}{{(x+\frac{3a-2}{6(2a-1)})}^2}-(x+a)ln(1+\frac{1}{x})$$

and

$$t(x)=1+\frac{(a-\frac{1}{2})x}{{(x+\frac{3a-2}{6(2a-1)}-\frac{7a^2-8a+2}{24{(2a-1)}^{2}x})}^2}-(x+a)ln(1+\frac{1}{x}).$$

We have

$$s^{″}(x)=-\frac{P(x)}{x^2{(x+1)}^2{(3a-6x+12ax-2)}^4},$$

(9)

where

$$\begin{array}{rcl}P(x)& =& 648(7a^2-8a+2){(2a-1)}^3{x}^3\\ +48(18a-11)(3a-2)(3a-1){(2a-1)}^2{x}^2\\ +(27a-2)(2a-1){(3a-2)}^{3}x\\ +a{(3a-2)}^4\end{array}$$

and

$$t^{″}(x)=\frac{Q(x)}{x^2{(x+1)}^2{(24{(2a-1)}^2{x}^2+4(2a-1)(3a-2)x-(7a^2-8a+2))}^4},$$

(10)

where

$$\begin{array}{rcl}Q(x)& =& 3,072(558a-1,098a^2+675a^3-92){(2a-1)}^6{x}^6\\ +64(94,140a^2-31,392a-122,256a^3+57,753a^4+3,860){(2a-1)}^5{x}^5\\ +128(816a-3,556a^2+7,164a^3-6,813a^4+2,484a^5-68){(2a-1)}^4{x}^4\\ +64(7a^2-8a+2)(942a^2-356a-1,086a^3+465a^4+48){(2a-1)}^3{x}^3\\ +16(10a-14a^2+9a^3-4){(7a^2-8a+2)}^2{(2a-1)}^2{x}^2\\ +(41a^2-24a-2){(7a^2-8a+2)}^3(2a-1)x\\ -a{(7a^2-8a+2)}^4.\end{array}$$

Proofs of Theorems 3 and 4 For $a\in [0,\frac{4-\sqrt{2}}{7})\cup (\frac{1}{2},\frac{4+\sqrt{2}}{7})$, the leading coefficients of the polynomials P and Q are negative. We are in a position to consider $j>0$ (depending on a) such that $P(x)<0$ and $Q(x)<0$, for every $x\in [j,\mathrm{\infty })$. By (9) and (10), s is concave and t is convex. But $s(\mathrm{\infty })=t(\mathrm{\infty })=0$, so $s<0$ and $t>0$ on $[j,\mathrm{\infty })$.

Now inequalities $s(x)<0$ and $t(x)>0$, for every $x\in [j,\mathrm{\infty })$ are (6) and (7) and we are done. □

Proof of Theorem 5 For $a\in (\theta ,1]$, the leading coefficients of polynomials P and Q are positive. We are in a position to consider $l>0$ (depending on a) such that $P(x)>0$ and $Q(x)>0$, for every $x\in [l,\mathrm{\infty })$. By (9) and (10), now s is convex and t is concave. But $s(\mathrm{\infty })=t(\mathrm{\infty })=0$, so $s>0$ and $t<0$ on $[l,\mathrm{\infty })$.

Now inequalities $s(x)>0$ and $t(x)<0$, for every $x\in [l,\mathrm{\infty })$ are (6) and (7) and we are done. □

The special case $a=1/2$ provides the approximation formula

$${(1+\frac{1}{n})}^{n+\frac{1}{2}}\approx exp(1+\frac{n}{12{(n+\frac{1}{3})}^3}).$$

It is sustained by the following.

Theorem 6 For every real number $x\ge 1$, the following inequalities hold:

$$exp(1+\frac{x}{12{(x+\frac{1}{3})}^3})<{(1+\frac{1}{x})}^{x+\frac{1}{2}}<exp(1+\frac{x}{12{(x+\frac{1}{3}-\frac{7}{90x})}^3}).$$

(11)

Proof Let us define the functions

$$b(x)=1+\frac{x}{12{(x+\frac{1}{3})}^3}-(x+\frac{1}{2})ln(1+\frac{1}{x})$$

and

$$c(x)=1+\frac{x}{12{(x+\frac{1}{3}-\frac{7}{90x})}^3}-(x+\frac{1}{2})ln(1+\frac{1}{x}).$$

We have

$$b^{″}(x)=-\frac{15x+171x^2+189x^3+1}{2x^2{(x+1)}^2{(3x+1)}^5}$$

and

$$c^{″}(x)=\frac{R(x-1)}{2x^2{(x+1)}^2{(30x+90x^2-7)}^5},$$

where

$$\begin{array}{rcl}R(x)& =& 44,117,710,950x+122,222,582,550x^2+187,066,692,000x^3\\ +170,643,105,000x^4+92,646,666,000x^5+27,671,382,000x^6\\ +3,499,200,000x^7+6,793,216,207.\end{array}$$

Evidently, b is concave, c is convex, with $b(\mathrm{\infty })=c(\mathrm{\infty })=0$, so $b<0$ and $c>0$ on $[1,\mathrm{\infty })$. The proof is completed. □

In case $a=1/2$, the entire asymptotic representation

$${(1+\frac{1}{x})}^{x+1/2}\sim exp(1+\frac{\frac{1}{12}x}{{(x+{\sum }_{j=0}^{\mathrm{\infty }}\frac{a_j}{x^j})}^3}),x\to \mathrm{\infty },$$

(12)

can be constructed. In this sense, we write (12) as

$${(f(x))}^{-1/3}\sim x+\sum _{j=0}^{\mathrm{\infty }}\frac{a_j}{x^j},$$

where

$$f(x)=\frac{(x+\frac{1}{2})ln(1+\frac{1}{x})-1}{\frac{1}{12}x}.$$

By using the Maclaurin expansion of $ln(1+t)$, with $t=x^{-1}$, we deduce that

$$f(x)=\sum _{j=3}^{\mathrm{\infty }}{(-1)}^{j-1}\frac{6(j-2)}{j(j-1)x^j}.$$

Now the coefficients $a_j$ in (12) can be inductively obtained by equating the following relation:

$$\left\{\sum _{j=3}^{\mathrm{\infty }}{(-1)}^{j-1}\frac{6(j-2)}{j(j-1)x^j}\right\}\times {\{x+\sum _{j=0}^{\mathrm{\infty }}\frac{a_j}{x^j}\}}^3=1.$$

The first coefficients are $a_0=\frac{1}{3}$, $a_1=-\frac{7}{90}$ (see (11)), then $a_2=\frac{16}{405}$, $a_3=-\frac{2,141}{85,050}$, ….

Further research in the problem of approximating the constant e can be found in [1–4].

Finally, we leave as an open problem the approximation formula (5) for values of $a\in [0,1]$ other than those discussed in this paper.