4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ You can use limits I think $\endgroup$ – Archis Welankar Aug 15 '16 at 17:47
  • $\begingroup$ @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! $\endgroup$ – Stupid_Guy Aug 15 '16 at 17:48
  • $\begingroup$ 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. $\endgroup$ – mathguy Aug 15 '16 at 18:22
6
$\begingroup$

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 $$\int_{1-1/n}^1 \frac 1t \, dt \le \frac{n}{n-1} \cdot \frac 1n = \frac 1{n-1}.$$

$\endgroup$
6
$\begingroup$

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, $$b_{n+1}^{-1/n}=\frac{1+(n-1)(1-\frac{1}{n})^{-1}}{n}\geq \left(\left(1-\frac{1}{n}\right)^{-(n-1)}\right)^{1/n}=b_{n}^{-1/n}.$$

$\endgroup$
  • $\begingroup$ Another way forward uses the more elementary Bernoulli's Inequality. I've posted a solution herein that uses that approach. $\endgroup$ – Mark Viola Aug 15 '16 at 19:14
2
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.