How to prove $\lim_{n\to\infty}\frac{1}{\Gamma(n/2+1)}\int_{0}^{n} t^{n/2}e^{-t}dt = 1$? How can we prove
$$\lim_{n\to\infty}\frac{1}{\Gamma(n/2+1)}\int_{0}^{n}  t^{n/2}e^{-t}dt = 1$$
? 
We ran experiments on MATLAB, it seems the statement is true.
 A: Since the maximum of $t^{n/2}e^{-t/2}=e^{-\frac t2+\frac n2\log(t)}$ occurs at $t=n$, we get that
$$
t^{n/2}e^{-t/2}\le n^{n/2}e^{-n/2}\tag1
$$
Therefore,
$$
\begin{align}
1-\frac1{\Gamma(n/2+1)}\int_0^n t^{n/2}e^{-t}\,\mathrm{d}t
&=\frac1{\Gamma(n/2+1)}\overbrace{\int_n^\infty t^{n/2}e^{-t}\,\mathrm{d}t}^{\Gamma(n/2+1,n)}\tag2\\
&\le\frac{\left(\frac ne\right)^{n/2}}{\Gamma(n/2+1)}\int_n^\infty e^{-t/2}\,\mathrm{d}t\tag3\\
&=\frac{\left(\frac ne\right)^{n/2}}{\Gamma(n/2+1)}2e^{-n/2}\tag4\\
&\le\frac{\left(\frac ne\right)^{n/2}}{\sqrt{\pi n}\left(\frac n{2e}\right)^{n/2}}2e^{-n/2}\tag5\\
&=\frac2{\sqrt{\pi n}}\left(\frac2e\right)^{n/2}\tag6
\end{align}
$$
Explanation:
$(2)$: $\Gamma(n/2+1)=\int_0^\infty t^{n/2}e^{-t}\,\mathrm{d}t$ is the Gamma Function
$\phantom{\text{(2):}}$ $\Gamma(n/2+1,n)=\int_n^\infty t^{n/2}e^{-t}\,\mathrm{d}t$ is the Incomplete Gamma Function
$(3)$: apply $(1)$
$(4)$: integrate
$(5)$: Stirling is an underapproximation (see Theorem $4$)
$(6)$: simplify
Thus,
$$
\bbox[5px,border:2px solid #C0A000]{1-\frac2{\sqrt{\pi n}}\left(\frac2e\right)^{n/2}\le\frac1{\Gamma(n/2+1)}\int_0^n t^{n/2}e^{-t}\,\mathrm{d}t\le1}\tag7
$$
Apply the Squeeze Theorem.
The bound in $(6)$ is pretty close as $n\to\infty$:

A: In terms of the normalised lower incomplete gamma function $P(a,z)$,
$$
\mathop {\lim }\limits_{n \to  + \infty } \frac{1}{\Gamma(n/2+1)}\int_0^n {t^{n/2} \mathrm{e}^{ - t} \mathrm{d}t}  = \mathop {\lim }\limits_{n \to  + \infty } P\!\left( {\tfrac{n}{2} + 1,n} \right) \le \mathop {\lim }\limits_{n \to  + \infty } (1 - \mathrm{e}^{ - n} )^{\frac{n}{2} + 1}  = 1,
$$
where I used http://dlmf.nist.gov/8.10.E11.
A: $$\dfrac{1}{\Gamma(n/2+1)} \int_{0}^n t^{n/2} e^{-t}\; dt = \mathbb P(X \le n)$$ where the random variable $X$ has a Gamma distribution with scale parameter $1$ and shape parameter $n/2+1$. This distribution has mean $n/2+1$ and standard deviation $\sqrt{n/2+1}$, so $n$ is approximately $\sqrt{n/2}$ standard deviations above the mean.
By Chebyshev's inequality, the probability goes to $1$ as $n \to \infty$.
A: $$\int t^{\frac n2}\,e^{-t}\,dt=-\Gamma \left(\frac{n}{2}+1,t\right)$$
$$\int_0^n t^{\frac n2}\,e^{-t}\,dt=\Gamma \left(\frac{n}{2}+1,0\right)-\Gamma \left(\frac{n}{2}+1,n\right)=\Gamma \left(\frac{n}{2}+1\right)-\Gamma \left(\frac{n}{2}+1,n\right)$$
$$\frac 1 {\Gamma \left(\frac{n}{2}+1\right)}\int_0^n t^{\frac n2}\,e^{-t}\,dt=1-\frac{\Gamma \left(\frac{n}{2}+1,n\right) } { \Gamma \left(\frac{n}{2}+1\right)}$$ The last term decreases very fast (have a look at dlmf for the asymptotics).
