Prove the inequality $\frac{|x|^n}{n!} \le e^x +e^{-x}$

Question

I'm trying to prove the following inequality: $$\frac{|x|^n}{n!} \le e^x+e^{-x}$$ for every $x \in \mathbb{R}$, using the Taylor series $$e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$$ If $x$ is positive, then $e^x \ge x^n/n!$ because the terms of the series are non negative, but this doesn't finish the proof. On the other hand I tried with the following passages: $$e^x+e^{-x}=\sum_{n=0}^\infty \left( \frac{x^n}{n!}+\frac{(-x)^n}{n!}\right)=\sum_{n=0}^\infty \frac{x^n+(-x)^n}{n!}=\sum_{n=0}^\infty \frac{x^n}{n!}(1-(-1)^n)$$ But this doesn't seem to go anywhere. Probably there is something trivial that I'm not seeing, but I don't know how to do it.

Answer 1

The inequality does not change if $x$ is substituted by $-x$, so proving it for $x \ge 0$ (as you did) is sufficient: $$ e^x + e^{-x} \ge e^x=\sum_{k=0}^\infty \frac{x^k}{k!} \ge \frac{x^n}{n!} = \frac{|x|^n}{n!} \, . $$

But of course you can do the same estimate for $x < 0$ explicitly: $$ e^x + e^{-x} \ge e^{-x}=\sum_{k=0}^\infty \frac{(-x)^k}{k!} = \sum_{k=0}^\infty \frac{|x|^k}{k!}\ge \frac{|x|^n}{n!} \, . $$

Answer 2

As an alternative using that $n!\ge\frac{n^n}{e^n}$, as proved for example here, we have

$$\frac{|x|^n}{n!} \le \left(\frac{e|x|}{n}\right)^n\le e^{|x|}\le e^x +e^{-x}$$

indeed

$$\left(\frac{e|x|}{n}\right)^n\le e^{|x|} \iff \frac{|x|}{n}\le e^{\frac{|x|}n-1}$$