Limit of an expression $$\lim\limits_{n\to\infty}\frac{1}{e^n\sqrt{n}}\sum\limits_{k=0}^{\infty}\frac{n^k}{k!}|n-k|=\sqrt{2/\pi}$$
Is this limit true? I should show limit is true. It is allowed to use computer programs to find this limit.
Thanks for your helps...
 A: This question has a nice probabilistic interpretation. Given that $X$ is a Poisson distribution with parameter $\lambda=n$, we are essentially computing the expected value of the absolute difference between $X$ and its mean $n$. The central limit theorem gives that $Y\sim N(n,n)$ (a normal distribution with mean and variance equal to $n$) is an excellent approximation of our distribution for large values of $n$, hence:
$$\begin{eqnarray*}\frac{1}{e^n \sqrt{n}}\sum_{k=0}^{+\infty}\frac{n^k}{k!}|n-k|&\approx&\frac{1}{\sqrt{n}}\cdot\frac{1}{\sqrt{2\pi n}}\int_{-\infty}^{+\infty}|x-n|\exp\left(-\frac{(x-n)^2}{2n}\right)\,dx\\&=&\frac{2}{n\sqrt{2\pi}}\int_{0}^{+\infty}x\exp\left(-\frac{x^2}{2n}\right)\,dx\\&=&\color{red}{\sqrt{\frac{2}{\pi}}},\end{eqnarray*}$$
so the limit is not zero.
A: Split the series at $k = n$:
\begin{align}
\sum_{k=0}^\infty \frac{n^k}{k!}\lvert n-k\rvert &= \sum_{k=0}^n \frac{n^k}{k!}(n-k) + \sum_{k=n+1}^\infty \frac{n^k}{k!}(k-n)\\
&= \sum_{k=0}^n\frac{n^{k+1}}{k!} - \sum_{k=1}^n \frac{n^k}{(k-1)!} + \sum_{k=n+1}^\infty \frac{n^k}{(k-1)!} - \sum_{k=n+1}^\infty \frac{n^{k+1}}{k!}\\
&= \sum_{k=0}^n \frac{n^{k+1}}{k!} - \sum_{k=0}^{n-1} \frac{n^{k+1}}{k!} + \sum_{k=n}^\infty \frac{n^{k+1}}{k!} - \sum_{k=n+1}^\infty \frac{n^{k+1}}{k!}\\
&= 2\frac{n^{n+1}}{n!}.
\end{align}
So you are looking for
$$\lim_{n\to\infty} \frac{2n^{n+\frac{1}{2}}}{n!e^n}.$$
Now recall or look up Stirling's formula.
