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Proving $(1 + 1/n)^{n+1} \gt e$

How to prove this:

$$ \left(\frac{x}{x-1}\right)^x \geq e \qquad\text{for}\qquad x \in \mathbb{N}^* $$

$e$ is the base of the natural logarithm.

and I think the equal satisfies if $x$ converges to infinity.

Thank you!

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marked as duplicate by Henry, Bruno Joyal, David Mitra, The Chaz 2.0, robjohn May 7 '12 at 21:08

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Related (or duplicate??) – The Chaz 2.0 May 7 '12 at 19:31
Possibly helpful: $$\left(\frac{x}{x-1}\right)^x=\left(\frac{x^2+x}{x^2-1}\right)^x=\left(\frac{x^‌​2}{x^2-1}\right)^x\left(\frac{x^2+x}{x^2}\right)^x=\left(\frac{x^2}{x^2-1}\right)‌​^x\left(1+\frac{1}{x}\right)^x$$ – Alex Becker May 7 '12 at 19:38
Thanks. I should search more thoroughly before posting. – Crend King May 7 '12 at 19:41
Yes, but only to a point. If the linked question were not in the list of my "answers", I might have never found it! – The Chaz 2.0 May 7 '12 at 19:43
Don't the other question deal with $n$ natural, while this deals with $x$ real? Of course, the answers there can probably be modified... – Aryabhata May 7 '12 at 19:53

2 Answers 2

Hint: Show that $\lim_{x\to\infty}f(x)=e$ and that $f'(x)\ne0$ (i.e. there are no stationary points)

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$f(x) = \frac{-1}{x}$ for $x \gt 0$. $\lim_{x \to \infty} f(x) = 0$, and $f'(x) \neq 0$. Does not imply $f(x) \ge 0$. Are you using some additional assumptions which you haven't mentioned? – Aryabhata May 7 '12 at 20:09
The hint leads to an argument that the graph of $f$ can't cross the line $y=e$. Checking $f(2)$ will help finish the problem. – Jonas Meyer May 7 '12 at 20:20

First off, $\frac{x}{x-1} > 0$ iff $x < 0$ or $x > 1$, so we can't take the natural logarithm if $x\in[0,1]$. My answer addresses the inequality for real-valued $x$, as in the original post (proving it for $x > 1$ and disproving it for $x < 0$). Now $$ e\le \left(\frac{x}{x-1}\right)^{x}= \left(1-\frac1x\right)^{-x} \to e $$ already guarantees the limiting behavior for us. Depending on the sign of $x$, this inequality becomes $$ e^{-1/x} \le 1-\frac1x \qquad(x < 0) $$ $$ e^{-1/x} \ge 1-\frac1x \qquad(x > 1) $$ Setting $t=\frac1x\lt1$ and using the Taylor series, this translates to $$ \eqalign{ 1-t &\le e^{-t} = \sum_{n=0}^\infty\frac{(-1)^n t^n}{n!} \\ &\le 1-t+\frac{t^2}2+\frac{t^3}6+\cdots } $$ for $t \in (0,1)$, which is patently true, while for negative values of $t$, the reversed inequality is patently false (which we can easily check in the original by trying $x=-1$ since $2 < e$). Therefore I would suggest adding the stipulation that $x > 0$ (necessarily) or actually, $x > 1$.

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$(1 - 1/x)^{-x} \to e^x$? – TMM May 7 '12 at 20:39
How can $x$ be in the limit after $x\to\infty$? – Dirk May 7 '12 at 20:57
@TMM: Yes, actually for $a\in\mathbb{C}$ and $b\in\mathbb{R}$ (as is easy to generalize from the case $a=b=1$), $$\lim_{x\to\infty}\left(1+\frac{a}{x}\right)^{bx}=e^{ab}\,.$$ I was using the special case $a=b=-1$. To get the general formula starting with $a=b=1$, first introduce the $b$ and use that the function $x\to x^b$ is continuous. Then to indroduce the $a$, use the substitution $x\leftarrow \frac{x}{a}$. – bgins May 7 '12 at 21:04
@bgins: You are missing the point. $(1 - 1/x)^{-x} \to e$ and not $e^x$. – TMM May 7 '12 at 23:09
@TMM (and Dirk): yes, of course. I got it now. – bgins May 8 '12 at 7:25

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