TITLE: Prove this inequality $1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots+\frac{x^n}{n!}>\frac{e^x}{2}$
QUESTION [14 upvotes]: let $0<x\le n$,prove that
$$1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\cdots+\dfrac{x^n}{n!}>\dfrac{e^x}{2}$$
Now I have prove $x=n$ case and my methods is very ugly,but $x\in(0,n]$,I can't prove it 
Thank you 
the following it's $x=n$ solution
show that
$$1+\dfrac{n}{1!}+\dfrac{n!}{2!}+\cdots+\dfrac{n^n}{n!}>\dfrac{e^n}{2},n\ge 0,n\in Z$$
use this 
$$e^n=\sum_{k=0}^{n}\dfrac{n^k}{k!}+\dfrac{1}{n!}\int_{0}^{n}(n-t)^ne^tdt$$
then
$$\Longleftrightarrow n!>2e^{-n}\int_{0}^{n}(n-t)^ne^tdt$$
$$\Longleftrightarrow \int_{0}^{\infty}t^ne^{-t}dt>2e^{-n}\int_{0}^{n}(n-t)^ne^tdt$$
let $u=n-t$ 
$$\Longleftrightarrow \int_{0}^{\infty}t^ne^{-t}dt>2\int_{0}^{n}u^ne^{-u}du$$
$$\Longleftrightarrow \int_{n}^{\infty}u^ne^{-u}du>\int_{0}^{n}u^ne^{-u}du$$
let $f(u)=u^ne^{-u}$
$$\Longleftrightarrow  f(n+h)\ge f(n-h),0\le h\le n$$
This is very easy.

REPLY [6 votes]: Here is a reduction to the case $x=n$.
Notice that by integration by parts we have
$$\int_x^\infty \frac{t^n}{n!} e^{-t} \, dt = e^{-x}\left(1 + x + \frac{x^2}{2!} + \dots + \frac{x^n}{n!}\right).$$
Thus it is enough to show that
$$\int_x^\infty \frac{t^n}{n!} e^{-t} \, dt > \frac{1}{2}.$$
But clearly $\int_x^\infty \frac{t^n}{n!} e^{-t} \, dt \ge \int_n^\infty \frac{t^n}{n!} e^{-t} \, dt$ when $0 < x \le n$.