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How to prove $(1+1/x)^x$ is increasing when $x>0$?

Let $k \in \mathbb{N}$ be fixed. For all $n \in \mathbb{N}$, is it true that

$$\left(\frac{n-1}{n}\right)^{n-1} \leq \left(\frac{n-k-1}{n-k}\right)^{n-k-1}?$$

Maple suggests the inequality holds, but I see no straightforward way to compare these two quantities; the fraction on the left is closer to one but has a higher exponent compared with the righthand side.

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I do not have an answer, but I considered this : you can reduce yourself to the case $k=1$ because the rest follows by induction. Is there anyway this is helping? Sunni's answer does the trick, but this is intriguing me ; I am wondering if there is a proof where you remain in the integers. –  Patrick Da Silva Dec 16 '11 at 0:55
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@cardinal: indeed, this inequality follows from the reciprocal of that inequality with $n-1$ here playing the role of $n$ there. –  robjohn Dec 16 '11 at 1:59
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marked as duplicate by Qiaochu Yuan Dec 18 '11 at 21:15

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3 Answers

up vote 6 down vote accepted

$\displaystyle\left(\frac{x-1}{x}\right)^{x-1}=\left(1-\frac{1}{x}\right)^{x-1}$ is a decreasing function since the derivative of it's logarithm is $$ \log\left(1-\frac{1}{x}\right)+\frac{1}{x}<0\tag{1} $$ for all $x>1$. This follows from the common inequality $$ e^x\ge1+x\tag{2} $$ for all $x$.

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Consider $f(x)=(a-x)[\log (a-x)-\log(a-x+1)]$. It is not difficult to show $f(x)$ is an increasing function over $x\in (0, a)$.

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If $n > k$ it's sufficient to prove that the function $f(x) = (\frac{x-1}{x})^{x-1}$ is decreasing for positive $x$. Rewrite it as $f(x) = e^{(x-1)\log(1-1/x)}$. Note that $\lim_{x\rightarrow\infty} (x-1)\log(1-\frac{1}{x}) = \lim_{x\rightarrow\infty} x\log(1-\frac{1}{x}) - \log(1-\frac{1}{x}) = \log e^{-1} - 0 = -1$. Furthermore, $(x-1)\log(1-1/x)$ is always nonpositive if $x$ is positive. So $f$ behaves like the exponential function over $(-1,0)$; it decreases.

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