Discussion of $f_n(x) = n\log(x) - x$ At first I had to show that $\lim_{n\rightarrow\infty} \frac{1}{n!} \int_{n}^{\infty}x^ne^{-x} dx = \frac{1}{2}$
Using stirlings formula and appropriate substitution I have shown the above. I now have to discuss the function $f_{n}(x) = n \log(x) - x, \;\;\forall n\in\mathbb{N}$. Using $f_n(x)$ I want to find useful estimations for:
$\int_{n}^{n+n^\alpha}x^ne^{-x} dx $ and $\int_{n+n^\alpha}^{\infty}x^ne^{-x} dx $ for an appropriate $\alpha <1 $
I can't figure out how the above defined $f_n(x)$ should help me to find proper estimations for the given integrals.
Could someone give me a hint?
That would be very nice!
Thomas
 A: We shall first examine the integral $I_n$, given by 
$$\begin{align}I_n&=\int_0^n x^ne^{-x}\,dx\\\\
&=\int_0^{n} e^{f_n(x)}\,dx \tag 1
\end{align}$$
where $f_n(x)=n\log(x)-x$.
Enforcing the substitution $x\to (n-x)/\sqrt{2n}$ in $(1)$, we find that 
$$\begin{align}
I_n&=\sqrt{2n}\int_0^{\sqrt{n/2}} e^{f_n\left(\frac{n-x}{\sqrt{2n}}\right)}\,dx\\\\
&=\sqrt{2n}\left(\frac ne\right)^n\int_0^{\sqrt{n/2}} e^{-x^2}e^{g_n(x)}\,dx\tag2
\end{align}$$
where $g_n(x)=n\left(\log\left(1-x/\sqrt{n/2}\right)+x/\sqrt{n/2}\right)+x^2$. 

At this point, we note that $g_n(x)\le 0$ for $x\in [0,\sqrt{n/2})$.  Therefore, $e^{g_n(x)}\le 1$ for $x\in [0,\sqrt{n/2})$.  Moreover, for $x\in (0,\sqrt{n/2})$
$$\lim_{n\to \infty}e^{g_n(x)}=1$$
Applying the dominated convergence theorem, we find 
$$\begin{align}
\lim_{n\to \infty} \int_0^{\sqrt{n/2}} e^{-x^2}e^{g_n(x)}\,dx&=\int_0^\infty e^{-x^2}\,\lim_{n\to \infty}\left(e^{g_n(x)}\xi_{[0,\sqrt{n/2}]}(x)\right)\,dx\\\\
&=\int_0^\infty e^{-x^2}\,dx\\\\
&=\sqrt{\pi}/2\tag 3
\end{align}$$
Divinding $(2)$ by $n!$, applyng Stirling's formula, and using $(3)$ yields the result reported in the OP
$$\lim_{n\to\infty}\frac1{n!}\int_0^n x^n e^{-x}\,dx=\frac12$$

MORE TO COME
