Series $\frac{k^2}{k!}$ with $ k=1$ to $\infty$ A practice Math Subject GRE asked me to compute
$\sum_{k=1}^\infty \frac{k^2}{k!}$. The sum is equal to $2e$, but I wasn't able to figure this out using Maclarin series or discrete PDFs. What's the most elementary way to solve this? Thank you.

$e^x=1+x+\dfrac{x^2}{2!}+\cdots+\dfrac{x^k}{k!}+\cdots$
Differentiating both sides,
$e^x=0+1+\dfrac{2x}{2!}+\cdots+\dfrac{kx^{k-1}}{k!}+\cdots$
Multiplying both sides by $x$,
$xe^x=x+\dfrac{2x^2}{2!}+\cdots+\dfrac{kx^k}{k!}+\cdots$
Again differentiating both sides,
$(x+1)e^x=1+\dfrac{2^2x}{2!}+\cdots+\dfrac{k^2x^{k-1}}{k!}+\cdots$
Now putting $x=1$ gives,
$2e=\dfrac{1^2}{1!}+\dfrac{2^2}{2!}+\cdots+\dfrac{k^2}{k!}+\cdots$ 
