Manipulating an infinite series I want to show (or rather, know why) $$\sum_{k = 1}^{\infty} \frac{k^2}{k!} = 2e.$$ I have tried using the power series expansion for $e^x$, and also series manipulations with $\ln(x)$, but with no success. Any hints would be appreciated.
 A: $$\sum_{k=1}^{+\infty}\frac{k^2}{k!}=\sum_{k=1}^{+\infty}\frac{k}{(k-1)!}=\sum_{k=1}^{+\infty}\frac{k-1}{(k-1)!}+\sum_{k\geq 1}\frac{1}{(k-1)!}=2\sum_{k\geq 0}\frac{1}{k!}=\color{red}{2e}.$$
An alternative approach is the following one. Since:
$$\sum_{k\geq 1}\frac{e^{kx}}{k!} = e^{e^x}-1, \tag{1} $$
we have:
$$ \frac{d^2}{dx^2}\sum_{k\geq 1}\frac{e^{kx}}{k!}=\sum_{k\geq 1}\frac{k^2 e^{kx}}{k!}=e^{x+e^x}+e^{2x+e^x}\tag{2}$$
and it is enough to replace $x$ with $0$.
A: Recall that the expansion for $e^x$ is given by 
$$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}\tag 1$$
Differentiating $(1)$ and mulitplying by $x$ yields
$$xe^x=\sum_{n=1}^{\infty}\frac{nx^{n}}{n!}\tag 2$$
Differentiating $(2)$ and multiplying by $x$ yields
$$x(x+1)e^x=\sum_{n=1}^{\infty}\frac{n^2x^{n}}{n!}\tag 3$$
whereupon setting $x=1$ reveals that 
$$\bbox[5px,border:2px solid #C0A000]{2e=\sum_{n=1}^{\infty}\frac{n^2}{n!}}$$
A: Hint: $\frac{k^2}{k!}=\frac{k}{(k-1)!}=\frac{1+(k-1)}{(k-1)!}$. You also need to pay attention at the beginning of the series as well.
