How to prove $\lim_{n\to\infty}\frac{1}{\Gamma(n)}\int_{n}^{\infty}t^{n-1}e^{-t}dt = 1/2$? How to prove 
$$
\lim_{n\to\infty}\frac{1}{\Gamma(n)}\int_{n}^{\infty}t^{n-1}e^{-t}dt = 1/2?
$$
This is confirmed numerically.

It is also stated in
Limit involving incomplete gamma function
that $\lim_{n\to\infty}\gamma(n,n)/\Gamma(n,n) = 1$ as a fact. So I believe the above statement is true. But I am still searching for a proof.
 A: Here is a proof using the central limit thm.  Let $X_1,X_2,\ldots,X_n,\ldots$ be i.i.d. random variables with the exponential distribution $\operatorname{Exp}(1)$.  Then $X_1+X_2+\ldots+X_n$ is a random variable with a Gamma distribution:
$$X_1+X_2+\ldots+X_n\sim \Gamma(n,1).$$
Thus
$$S_n=\frac{X_1+X_2+\ldots+X_n-n}{\sqrt{n}}$$
tends to a standard Gaussian random variable $N$ (by the central limit thm).  That is,
$$\lim_{n\to\infty}\Bbb P[X_1+X_2+\ldots+X_n\ge n]=\lim_{n\to\infty}\Bbb P[S_n\ge 0]=\Bbb P[N\ge0]=\frac12.$$
Because
$$\Bbb P[X_1+X_2+\ldots+X_n\ge n]=\frac{1}{\Gamma(n)}\int_n^\infty t^{n-1}e^{-t}dt,$$
the claim follows.
A: Incomplete Gamma function has the following integral representation
$$\Gamma(a,z)=\int_z^\infty t^{a-1}e^{-t}dt$$
and then your limit can be written as
$$\lim_{n\to\infty}\frac{1}{\Gamma(n)}\int_{n}^{\infty}t^{n-1}e^{-t}dt=\lim_{n\to\infty}\frac{\Gamma(n,n)}{\Gamma(n)}$$
and using the asymptotic behaviour of each one (see incomplete Gamma)
$$\Gamma(n,n)\sim n^{n-1}e^{-n}\sqrt{\pi/2}\,n^{1/2}$$
and (see Gamma)
$$\Gamma(n)\sim n^{n-1}e^{-n}\sqrt{2\,\pi}\,n^{1/2}$$
you get your desired result.
A: Let $ n $ be a positive integer.
\begin{aligned}\small\frac{1}{\Gamma\left(n+1\right)}\int_{n+1}^{+\infty}{x^{n}\mathrm{e}^{-x}\,\mathrm{d}x}&\small=\frac{1}{\Gamma\left(n+1\right)}\int_{0}^{+\infty}{x^{n}\mathrm{e}^{-x}\,\mathrm{d}x}-\frac{1}{\Gamma\left(n+1\right)}\int_{n}^{n+1}{x^{n}\mathrm{e}^{-x}\,\mathrm{d}x}-\frac{1}{\Gamma\left(n+1\right)}\int_{0}^{n}{x^{n}\mathrm{e}^{-x}\,\mathrm{d}x}\\&\small=1-\frac{\mathrm{e}^{-n}}{n!}\int_{0}^{1}{\left(n+x\right)^{n}\mathrm{e}^{-x}\,\mathrm{d}x}-\frac{\sqrt{n}\mathrm{e}^{-n}}{n!}\int_{0}^{\sqrt{n}}{\left(n-\sqrt{n}y\right)^{n}\mathrm{e}^{\sqrt{n}y}\,\mathrm{d}y}\\ &\small=1-\frac{n^{n}\mathrm{e}^{-n}}{n!}\int_{0}^{1}{\left(1+\frac{x}{n}\right)^{n}\mathrm{e}^{-x}\,\mathrm{d}x}-\frac{\sqrt{n}n^{n}\mathrm{e}^{-n}}{n!}\int_{0}^{\sqrt{n}}{\left(1-\frac{x}{\sqrt{n}}\right)^{n}\mathrm{e}^{\sqrt{x}}\,\mathrm{d}x}\end{aligned}
Notice that in the second line we substituted $ x=n-\sqrt{n}y \cdot $
Using Stirling's formula, we have that $ \frac{n^{n}\,\mathrm{e}^{-n}\sqrt{n}}{n!}\underset{n\to +\infty}{\sim}\frac{1}{\sqrt{2\pi}} $, and notice that $ \lim\limits_{n\to +\infty}{\left(1-\frac{x}{\sqrt{n}}\right)^{n}\mathrm{e}^{\sqrt{n}x}}=\mathrm{e}^{-\frac{x^{2}}{2}} $, for any positive real $ x \cdot $ So by applying the dominated convergence theorem on the function $ f_{n} $ defined on $ \mathbb{R}_{+} $ as follows : $$ \left(\forall x\in\mathbb{R}_{+}\right),\ f_{n}\left(x\right)=\left\lbrace\begin{aligned}\left(1-\frac{x}{\sqrt{n}}\right)^{n}\mathrm{e}^{\sqrt{n}x}\ \ \ \ \ \ \ & \textrm{If }0\leq x\leq\sqrt{n}\\ 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ &\textrm{If }\ \ \ \ x\geq \sqrt{n}\end{aligned}\right. $$We get that : $$ \small\lim_{n\to +\infty}{\int_{0}^{\sqrt{n}}{\left(1-\frac{x}{\sqrt{n}}\right)^{n}\mathrm{e}^{\sqrt{n}x}\,\mathrm{d}x}}=\lim_{n\to +\infty}{\int_{0}^{+\infty}{f_{n}\left(x\right)\mathrm{d}x}}=\int_{0}^{+\infty}{\lim_{n\to +\infty}{f_{n}\left(x\right)}\,\mathrm{d}x}=\int_{0}^{+\infty}{\mathrm{e}^{-\frac{x^{2}}{2}}\,\mathrm{d}x} $$
And since for every positive real $ x $, $ \left(1+\frac{x}{n}\right)^{n}\leq\mathrm{e}^{x} $, we get that : $$ \frac{n^{n}\mathrm{e}^{-n}}{n!}\int_{0}^{1}{\left(1+\frac{x}{n}\right)^{n}\mathrm{e}^{-x}\,\mathrm{d}x}\leq\frac{n^{n}\mathrm{e}^{-n}}{n!}\underset{n\to +\infty}{\sim}\frac{1}{\sqrt{2\pi n}}\underset{n\to +\infty}{\longrightarrow}0 $$
Hence : $$ \lim_{n\to +\infty}{\frac{1}{\Gamma\left(n+1\right)}\int_{n+1}^{+\infty}{x^{n}\mathrm{e}^{-x}\,\mathrm{d}x}}=1-\frac{1}{\sqrt{2\pi}}\int_{0}^{+\infty}{\mathrm{e}^{-\frac{x^{2}}{2}}\,\mathrm{d}x}=\frac{1}{2} $$
