How to prove that: $\lim_{n\to\infty}\frac1{n!}\int_0^ne^{-t}t^n\,dt.=\frac{1}{2}$ I want to Show that $$\lim_{n\to\infty}\frac1{n!}\int_0^ne^{-t}t^n\,dt=\frac12.$$
I tried to squeeze the sequence as follows
$$\frac{n^{n}(1-e^{-n})}{n!}= \frac1{n!}\int_0^ne^{-t}n^n\,dt\ge \frac1{n!}\int_0^ne^{-t}t^n\,dt \ge \frac1{n!}\int_0^ne^{-n}t^n\,dt =\frac{n^{n+1}e^{-n}}{(n+1)!} .$$
Then what next? any idea?
 A: We first give the integral in terms of Gamma and Incomplete Gamma function:
\begin{align}
\int^{n}_0 t^n e^{-t}\,dt =\int^{\infty}_0 t^n e^{-t}\,dt -\int^\infty_n t^n e^{-t}\,dt = \Gamma(n+1)-\Gamma(n+1,n) = n!-\Gamma(n+1,n)
\end{align}
Hence:
\begin{align}
\frac{1}{n!}\int^{n}_0 t^n e^{-t}\,dt=1-\frac{\Gamma(n+1,n)}{n!}
\end{align}
Now we use the result found in this answer (it is after edit II before the conclusion), namely:
\begin{align}
\lim_{n\to\infty}\frac{\Gamma(n+1,n)}{n!} = \frac{1}{2}
\end{align}
And finally we get:
\begin{align}
\lim_{n\to\infty}\frac{1}{n!}\int^{n}_0 t^n e^{-t}\,dt = 1-\frac{1}{2}=\frac{1}{2}
\end{align}

Edit (Alternative approach).
I added an extra approach that relies on the CLT. First let's split up the integral:
\begin{align}
\int^n_0 t^ne^{-t}\,dt =\int^{n+1}_0 t^ne^{-t}\,dt - \int^{n+1}_n t^ne^{-t}\,dt
\end{align}
Then we know from this answer the following:
\begin{align}\tag{1}
\lim_{n\to\infty} \frac{1}{n!}\int^{n+1}_0 t^ne^{-t}\,dt = \lim_{n\to\infty}\frac{1}{(n-1)!}\int^{n}_0 t^{n-1}e^{-t}\,dt  =\frac{1}{2}
\end{align}
For the other part, note that:
\begin{align}
0\leq \frac{1}{n!}\int^{n+1}_n t^n e^{-t}\,dt \leq \frac{1}{n!}e^{-n}\int^{n+1}_n t^n \,dt = \frac{1}{(n+1)!}e^{-n}\left((n+1)^{n+1}-n^{n+1} \right)
\end{align}
Using Stirling's asymptotics for the factorial we get:
\begin{align}
\lim_{n\to\infty}\frac{1}{(n+1)!}e^{-n}\left((n+1)^{n+1}-n^{n+1} \right) &=\lim_{n\to\infty} \frac{e^{-n}\left((n+1)^{n+1}-n^{n+1} \right)}{\sqrt[]{2\pi (n+1)}(n+1)^{n+1}e^{-(n+1)}}\\
&=\lim_{n\to\infty} \frac{e}{\sqrt[]{2\pi(n+1)}}\left(1-\left(1-\frac{1}{n+1} \right)^{n+1} \right)\\
&= 0
\end{align}
Using the squeeze theorem we get:
\begin{align}\tag{2}
\lim_{n\to\infty} \frac{1}{n!}\int^{n+1}_n t^n e^{-t}\,dt =0
\end{align} 
Putting (1) and (2) together gives the stated result, namely:
\begin{align}
\lim_{n\to\infty} \frac{1}{n!}\int^n_0 t^ne^{-t}\,dt = \lim_{n\to\infty} \frac{1}{n!}\int^{n+1}_0 t^ne^{-t}\,dt - \frac{1}{n!}\int^{n+1}_n t^ne^{-t}\,dt = \frac{1}{2}-0 =\frac{1}{2}
\end{align}
A: The same idea for proving
$$ \lim_{n\to +\infty}\frac{1}{e^n}\sum_{k=0}^{n}\frac{n^k}{k!}=\frac{1}{2}\tag{A}$$
applies here, too. The argument of the limit in the LHS of $(A)$ is the probability that a Poisson random variable takes a value which is $\leq $ than its expected value. The argument of your limit is the probability that a random variable with a $\Gamma$ distribution takes a value which is $\leq$ than its expected value. By the central limit theorem both limits equal $\frac{1}{2}$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\bbox[10px,#ffe]{\ds{\lim_{n \to \infty}\pars{{1 \over n!}\int_{0}^{n}\expo{-t}t^{n}\,\dd t}}}  =
\lim_{n \to \infty}\bracks{%
{1 \over n!}\int_{0}^{n}\expo{-\pars{n - t}}\pars{n - t}^{n}\,\dd t}
\\[5mm] = &\
\lim_{n \to \infty}\bracks{%
{\expo{-n}n^{n} \over n!}\int_{0}^{n}\expo{t}\pars{1 - {t \over n}}^{n}\,\dd t} =
\lim_{n \to \infty}\bracks{%
{\expo{-n}n^{n} \over n!}\int_{0}^{n}
\exp\pars{t + n\ln\pars{1 - {t \over n}}}\,\dd t}
\end{align}

At this point, I'll use the
  Laplace Method:

\begin{align}
&\bbox[10px,#ffe]{\ds{\lim_{n \to \infty}\pars{{1 \over n!}\int_{0}^{n}\expo{-t}t^{n}\,\dd t}}} =
\lim_{n \to \infty}\bracks{%
{\expo{-n}n^{n} \over n!}\int_{0}^{\infty}
\exp\pars{-\,{t^{2} \over 2n}}\,\dd t}
\\[5mm] & =
\lim_{n \to \infty}\bracks{%
{\expo{-n}n^{n} \over n!}\,\pars{\root{\pi \over 2}\,n^{1/2}}} =
{\root{2\pi} \over 2}\lim_{n \to \infty}\pars{%
{\expo{-n}n^{n + 1/2} \over n!}}
\\[5mm] & =
{\root{2\pi} \over 2}\lim_{n \to \infty}\pars{%
{\expo{-n}n^{n + 1/2} \over \root{2\pi}n^{n + 1/2}\expo{-n}}}\qquad
\pars{~n!\!-\!Stirling\ Asymptotic\ Expansion ~}
\\[5mm] & = \bbx{1 \over 2}
\end{align}
