How to prove this inequality or give a more accurate bound? How can we prove this inequality  or give a more accurate bound?
$$
1 + x + \frac{{{x^2}}}{{2!}} + ....... + \frac{{{x^n}}}{{n!}} > \frac{{{e^x}}}{2}，x \in [0,n]\
$$
I came across the equation:
$$
\mathop {\lim }\limits_{n \to \infty } \frac{{1 + n{\rm{ + }}\frac{{{n^2}}}{{2!}} + ..... + \frac{{{n^n}}}{{n!}}}}{{{e^n}}} = \frac{1}{2} 
$$
Maybe I can use this this inequality to solve this problem
 A: If $X$ is a random variable having Poisson distribution with parameter $x$, 
$e^{-x} \sum_{j=0}^n x^j/j! = {\mathbb P}(X \le n)$.  Thus your claim is that ${\mathbb P}(X \le n) \ge 1/2$ for $0 \le x \le n$.  Now ${\mathbb P}(X \le n)$ is a decreasing function of $x$ (e.g. because a Poisson random variable of parameter $x + \delta$, $\delta > 0$, can be realized as the sum of independent Poisson random variables of parameters $x$ and $\delta$), so it suffices to prove for the case $x=n$.  We can translate this to a question about a Poisson process with rate $1$: the number of arrivals in time interval $[0,n]$ has Poisson distribution with parameter $n$, and this $\ge n$ iff the $n$'th arrival occurs by time $n$. The time until the $n$'th arrival has a Gamma distribution.  If I'm not mistaken, Chen and Rubin showed that the median of this Gamma distribution is between $n-1/3$ and $n$, so in particular $P(X \ge n) > 1/2$.
A: Robert Israel already answered, below is a short self-contained argument.
We have $$\left(1+x+\frac{x^2}{2!}+\dots+\frac{x^n}{n!}\right)e^{-x}=\frac{1}{n!}\int_x^{\infty}e^{-t}t^ndt.$$
We immediately see that it suffices to consider $x=n$, and, taking into account that $n!=\int_{0}^{\infty} e^{-t}t^ndt$, we may rewrite our inequality as 
$$
\int_n^{\infty}e^{-t}t^ndt>\int_0^{n}e^{-t}t^ndt,
$$
denoting $t=ns$ we rewrite this as
$$
\int_1^{\infty}e^{-ns}s^nds>\int_0^{1}e^{-ns}s^nds.
$$
I claim that $$e^{-(1+x)}(1+x)>e^{-(1-x)}(1-x)\,\,\forall x\in (0,1).\,\,(*)$$  Taking $n$-th power of $(*)$ and integrating over $[0,1]$ we get that even
$$
\int_1^{2}e^{-ns}s^nds>\int_0^{1}e^{-ns}s^nds.
$$
For proving $(*)$ we rewrite it as $(1+x)/(1-x)>e^{2x}$ and expand both parts as power series of $x$: $$1+2x+2x^2+2x^3\dots>1+2x+2x^2+(2^3/3!)x^3+(2^4/4!)x^4+\dots,$$ this is clear.
