Prove this inequality $1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots+\frac{x^n}{n!}>\frac{e^x}{2}$

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.

Solution 1:

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 $0 < x \le n$.