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How to show $(1-\frac{1}{n})^n \leq \frac{1}{e} \leq(1-\frac{1}{n})^{n-1}$?

I can prove the first inequality: take the logarithm of both sides and then use the fact that $\log(1+x) \leq x.$

But how to prove the second inequality? The same method does not work here because we need an lower bound for $\log(1+x)$ now.

zxzx179
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  • You can use limits I think – Archis Welankar Aug 15 '16 at 17:47
  • @ArchisWelankar I guess what you mean is that the third term is decreasing and converges to $1/e$. I was wondering if there are some other proofs that do not involve limit.. Thanks so much! – zxzx179 Aug 15 '16 at 17:48
  • What definition of $e$ are you asked to use in this problem? For example, it may be easy to show that the first sequence is increasing and the second is decreasing; are you allowed to use the fact that both have the limit $1/e$? In that case, you are done. So, how you solve the problem depends on what definition of $e$ you are allowed to refer to. –  Aug 15 '16 at 18:22

3 Answers3

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Take logarithms in the second inequality to get $$-1 \le (n-1) \log(1-1/n)$$ which rearranges to $$- \log(1-1/n) \le \frac{1}{n-1}.$$ You can write this as $$\int_{1-1/n}^1 \frac 1t \, dt \le \frac{1}{n-1}.$$

Since $f(t) = \frac 1t$ is decreasing on $[1-1/n,1]$ its maximum value there is $1/(1-1/n) = n/(n-1)$. Consequently $$\int_{1-1/n}^1 \frac 1t \, dt \le \frac{n}{n-1} \cdot \frac 1n = \frac 1{n-1}.$$

Umberto P.
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It suffices to show that $a_n=(1-\frac{1}{n})^n$ is increasing and $b_n=(1-\frac{1}{n})^{n-1}$ is decreasing.

In order to see that $a_n$ is increasing consider the numbers $1$ and $n$ copies of $(1-\frac{1}{n})$, then by AGM inequality, $$a_{n+1}^{1/(n+1)}=\frac{1+n(1-\frac{1}{n})}{n+1}\geq \left(\left(1-\frac{1}{n}\right)^n\right)^{1/(n+1)}=a_{n}^{1/(n+1)}.$$

A similar strategy can be used for the sequence $b_n$. Consider the numbers $1$ and $n-1$ copies of $(1-\frac{1}{n})^{-1}$ , then by AGM inequality, $$b_{n+1}^{-1/n}=\frac{1+(n-1)(1-\frac{1}{n})^{-1}}{n}\geq \left(\left(1-\frac{1}{n}\right)^{-(n-1)}\right)^{1/n}=b_{n}^{-1/n}.$$

Robert Z
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  • Another way forward uses the more elementary Bernoulli's Inequality. I've posted a solution herein that uses that approach. – Mark Viola Aug 15 '16 at 19:14
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See THIS ANSWER for a more general development of the one we present herein.

We will show that $\left(1-\frac1n\right)^n$ and $\left(1-\frac1n\right)^{n-1}$ are increasing and decreasing sequences, respectively, using Bernoulli's Inequality.

Let $a_n=\left(1-\frac1n\right)^n$. Then, we have

$$\begin{align} \frac{a_{n+1}}{a_{n}}&=\frac{\left(1-\frac1{n+1}\right)^{n+1}}{\left(1-\frac1n\right)^n}\\\\ &=\left(1-\frac1n\right)\,\left(1+\frac{1}{n^2-1}\right)^{n+1} \tag 1\\\\ &\ge \left(1-\frac1n\right)\,\left(1+\frac{1}{n-1}\right) \tag 2\\\\ &=1 \end{align}$$

where in going from $(1)$ to $(2)$, we exploited Bernoulli's Inequality. Therefore, $a_n$ is monotonically increasing. Since its limit is $1/e$ we have

$$\bbox[5px,border:2px solid #C0A000]{\left(1-\frac1n\right)^n\le \frac1e} \tag 3$$

Let $b_n=\left(1-\frac1n\right)^{n-1}$. Then, we have

$$\begin{align} \frac{b_{n}}{b_{n+1}}&=\frac{\left(1-\frac1n\right)^{n-1}}{\left(1-\frac1{n+1}\right)^{n}}\\\\ &=\left(\frac{1}{\left(1-\frac1n\right)}\right)\left(1-\frac{1}{n^2}\right)^n \tag 4\\\\ &\ge \left(\frac{1}{\left(1-\frac1n\right)}\right)\left(1-\frac{1}{n}\right) \tag 5\\\\ &=1 \end{align}$$

where in going from $(4)$ to $(5)$, we exploited Bernoulli's Inequality again. Therefore, $b_n$ is monotonically increasing. Since its limit is $1/e$ we have

$$\bbox[5px,border:2px solid #C0A000]{\left(1-\frac1n\right)^{n-1}\ge \frac1e} \tag 6$$

Putting $(3)$ and $(6)$ together yields the coveted inequalities

$$\bbox[5px,border:2px solid #C0A000]{\left(1-\frac1n\right)^{n}\le \frac1e \le \left(1-\frac1n\right)^{n-1}}$$

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Mark Viola
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