How to prove $(1+\frac{1}{k})^k < e < (1+\frac{1}{k})^{k+1}$?

Question

I understand $(1+\frac{1}{k})^k$ converges as $e$ when $k$ goes to infinity. However, how to prove the inequality above? Please give me some hints.

Answer 1

Let $f$ be the function defined for $x>-1$ by $f(x)=\ln(1+x)$, and let $a$ be a positive real number. Using the Mean Value Theorem, there is a real number $\xi\in(0,a)$ such that $$\frac{\ln(1+a)}{a}=\frac{f(a)-f(0)}{a}=f'(\xi)=\frac{1}{1+\xi}\in\left(\frac{1}{1+a},1\right).$$ that is, for $a>0$, we have $$\frac{a}{1+a}<\ln(1+a)<a.$$ Applying the upper inequality with $a=1/n$, we get $$ n\ln\left(1+\frac{1}{n}\right)<1<(n+1)\ln\left(1+\frac{1}{n}\right)$$ Taking exponentials yields the desired inequality.$\qquad\square$

Answer 2

This solution is for the case when you are allowed to use the limit. Denote $a_k = (1+\frac{1}{k})^k$ and $b_k = (1+\frac{1}{k})^{k+1}$. First, show that the base case: $a_1 <e <b_1$. Second step, look at the ratios $\frac{a_{k+1}}{a_k}$ and $\frac{b_{k+1}}{b_k}$. Show that they monotone (a) increase and (b) decrease. Hence, draw your conclusion.

Answer 3

HINT:

1. Let's prove first that

$$(1+\frac{x}{m})^m< (1+\frac{x}{n})^n$$

if $x>0$, and $1\le m < n$ integers.

Indeed

$$(1+\frac{x}{m})^m = \sum_{k\ge 0} \frac{m(m-1)\cdots (m-k+1)}{m^k k!} x^k= \sum_{k \ge 0} (1-\frac{1}{m}) \cdots (1-\frac{k-1}{m}) \frac{x^k}{k!}$$

and each term is increasing with $m$.

2. From 1. we get $\left(1+\frac{x}{n}\right)^{\frac{n}{m}} > 1 + \frac{x}{m}$

that is

$$(1+ t)^a > 1 + a t$$

if $t> 0$ and $a>1$ is rational. Taking limits we get the inequality valid for all $a>1$. ( Bernoulli inequality). However, apriori for $a$ irrational it may not be strict. However, we can use the strict one for $a$ rational to prove the strict one as follows: write any $a>1$ as $a= a_1 \cdot a_2$, with $a_1$, $a_2>1$ and $a_1$ rational, and play with that

3. Let's show that

$$\left(1+\frac{1}{m}\right)^{m} < \left(1+\frac{1}{n}\right)^{n}$$

if $0< m < n$ ( not necessarily integers). Indeed, we have

$$(1+1/n)^{\frac{n}{m}} > 1+ \frac{n}{m}\cdot \frac{1}{n} = 1+ \frac{1}{m}$$ by Bernoulli.

4. Let us show that

$$(1+ \frac{1}{m})^{m+1}> (1+\frac{1}{m+\delta})^{m+ \delta + 1}$$

if $m> 0$, and $0< \delta \le 1$. Indeed, we have

$$\left(\frac{1+1/m}{1+1/(m+\delta)}\right)^{m+1} > 1+ (m+1)( \frac{1+1/m}{1+1/(m+\delta)}-1)$$

while

$$(1+ \frac{1}{m+\delta})^\delta \le 1+ \frac{\delta}{m+\delta}$$

since $\delta \le 1$ ( by Bernoulli, reversed). Now one checks that

$$1+ (m+1)\left( \frac{1+1/m}{1+1/(m+\delta)}-1\right)- (1+ \frac{\delta}{m+\delta}) = \frac{d^2}{m(m+\delta)(m+\delta+1)}>0$$

5. From 4. we get

$$(1+ \frac{1}{m})^{m+1} >(1+ \frac{1}{n})^{n+1}$$

if 0 < m < n

6.

$$(1+\frac{1}{n})^n < e < (1+\frac{1}{n})^{n+1}$$

for all $n> 0$ ( see graphs).

7.

The Bernoulli inequality $(1+x)^a > 1 + a x $, for $a>1$ holds if $x> -1$, $x\ne 0$. So one could try the above trick with series from 1. So it seems useful to note the power series expansion

$$(1-a x) ^{-1/a} = 1 + x + (1+a) \frac{x^2}{2} + (1+a)(1+2a)\frac{x^3}{3!} + (1+a)(1+2a)(1+3a) \frac{x^4}{4!} + \cdots $$

where the coefficients increase with positive $a$.

Therefore, all can be proved using Bernoulli's inequality. That one can be seen as: for $a> 1$ the function $t\mapsto t^a$ is strictly convex, so the linear estimate at $t=1$ is less than the function.

Answer 4

We have that

$$\left(1+\frac{1}{k}\right)^k < e < \left(1+\frac{1}{k}\right)^{k+1} \iff k\log\left(1+\frac{1}{k}\right)<1<(k+1)\log\left(1+\frac{1}{k}\right)$$

then, since for $x>0$ we have $\log(1+x)<x$, for the LHS inequality we have that

$$k\log\left(1+\frac{1}{k}\right) <k \cdot \frac 1k=1 $$

For the RHS inequality we have that for $x>0$

$$f(x)=(1+x)\log\left(1+\frac 1x\right) \implies f'(x)=\log\left(1+\frac 1x\right)-\frac1x<0$$

with $f(x)\to 1$ then

$$(k+1)\log\left(1+\frac{1}{k}\right)>1 $$