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I was going through my Lebesgue Measure and Integration course, and I came across this inequality. $$ \left(1+\frac{x}{n}\right)^{-n}\leq e^{-x} $$
I tried expanding both sides, and got
$$\text{LHS}= 1-x+\frac{n+1}{n}\frac{x^2}{2!}-\frac{(n+1)(n+2)}{n^2}\frac{x^3}{3!}\cdots $$
and $$ \text{RHS}=1-x+\frac{x^2}{2!}\cdots $$ Now, I noticed that all the terms in the LHS are either equal or bigger than the corresponding terms in RHS but I don't know how to conclude the inequality from this. Am I missing something extremely simple and obvious here?
Any help will be appreciated.

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    I'd recommend taking the derivative of $(1+x/n)^n - e^x$ after having taken inverses on both sides (be careful if you do take inverses of both sides) – 2017-02-06
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    The inequality is in reverse and holds for non-negative values of $x$ and positive integer values of $n$. – 2017-02-06
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    @ParamanandSingh Forgot to put in that $x$ is non-negative. – 2017-02-06
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    Sometimes an author might say something is "obvious" merely because it follows immediately from something that the author assumes the reader knows. – 2017-02-06

2 Answers 2

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For your inequality to be true, we need $n<0$ and $x \ge 0$. Note that from this link we know that $$e^{\frac{x}{n}} \ge \;1+\frac{x}{n}$$ Now, as $t^{-n}$ is an increasing function if $n<0$, $t>0$, we exponentiate each side by $-n$ to get $$e^{-x} \ge \bigg{(}1+\frac{x}{n}\bigg{)}^{-n}$$ as desired. Of course, if $n \ge 0$, the inequality signs reverse, for a similar reason. So your inequality is, for example, not true when $n=1$ as stated previously.

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    Let me see if there's any restriction.. – 2017-02-06
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    @Sum-Meister Okay. I will edit accordingly if so. – 2017-02-06
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    Rearranging the original inequality, we get: $e^x \leq (1+\frac{x}{n})^n$, which is, as we all know, false for all n, since this series is increasing and converges to $e^x$. – 2017-02-06
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    The other direction is true though. – 2017-02-06
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    @Omry The inequality in the OP is true if $n=-1$. – 2017-02-06
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    I was assuming n was natural. If it isn't, then substitute n->-n, and get $e^{-x}$'s series, which is decreasing and converges to it. – 2017-02-06
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    @Omry Well that wasn't stated, so I editted in a proof. – 2017-02-06
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    @Sum-Meister How long is this going to take? Not trying to hurry you on, but it does seem to be taking some time. Just going to ask when. – 2017-02-06
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We can take a logarithm from both sides. Then, we have $-n \log(1 + \frac{x}{n}) \leq -x \Rightarrow n \log(1 + \frac{x}{n}) \geq x \Rightarrow \log(1 + \frac{x}{n}) \geq \frac{x}{n}$

let $\frac x n = y$. Then, we have

$\log(1+y) \geq y$ and this is not true for any $y$