Showing that $\sum k^{2}/k!=e^2$ I'm having trouble seeing why $\sum_{k=1} k^{2}/k!=e^2$. The ratio test says that it converges absolutely. Pardon my ignorance, but are there any techniques to show this?
I thought about expanding $e^{2}= e \cdot e =(\sum_{k=0} \frac{1}{k!})(\sum_{k=0} \frac{1}{k!})$ but so far that hasn't helped me much.
 A: You may have misread your source, this is not equal to $e^2$, but to $2e$. Unrolling the expression,
$$\begin{align}
 \sum_{k=0}^\infty \frac{k^2}{k!} 
&= \sum_{k=1}^\infty \frac{k^2}{k!} 
= \sum_{k=1}^\infty \frac{k}{(k-1)!} 
= \sum_{k=0}^\infty \frac{k+1}{k!} = \sum_{k=0}^\infty \frac{k}{k!} +\sum_{k=0}^\infty \frac{1}{k!} \\
&
= \sum_{k=1}^\infty \frac{k}{k!} +e^1 
= \sum_{k=1}^\infty \frac{1}{(k-1)!} +e^1 
= \sum_{k=0}^\infty \frac{1}{k!} +e^1 
= e^1  +e^1 \\
&= 2e
\end{align}$$
A: \begin{align*}
\sum_{k=0}^\infty  \frac{k^2}{k!}&=\sum_{k=1}^\infty  \frac{k}{(k-1)!}
=\sum_{k=1}^\infty  \frac{k-1}{(k-1)!}+\sum_{k=1}^\infty  \frac{1}{(k-1)!}
\\ &=
\sum_{k=2}^\infty  \frac{k-1}{(k-1)!}+\sum_{k=0}^\infty  \frac{1}{k!}=\sum_{k=2}^\infty  \frac{1}{(k-2)!}+e 
=\sum_{k=0}^\infty  \frac{1}{k!}+e =2e
\end{align*}
A: $ e^x= \sum \frac{x^k}{k!} $
$ e^x= (e^x)'= \sum \frac{x^{(k-1)}}{(k-1)!} $
$ x e^x = \sum \frac{x^k}{(k-1)!} $
$ (x e^x)' = (x+1)e^x = \sum \frac{k x^{(k-1)}}{(k-1)!} = \sum \frac{k^2  x^{(k-1)}}{k!} $
so $2e = \sum \frac{k^2}{k!}$
A: No. Actually,
$\sum_{k=0} 2^{k}/k!=e^2$
