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.
 A: Consider the power series
$$f(x) = \sum_{k=0}^{\infty} \frac{x^k}{k!} = e^x$$
$$\frac{d}{dx} \left [x \frac{d}{dx} f(x)\right] = \sum_{k=0}^{\infty} \frac{k^2 x^k}{k!} = (x+1) e^x $$
Plug in $x=1$ and the stated result follows.
A: I'll try to calculate the more general series:
$$S_n=\sum_{k=0}^{\infty}\frac{k^n}{k!}$$ where $n$ is an integer number and $n\ge 0$.
\begin{align}
\begin{aligned}
S_n &=\sum_{k=0}^{\infty}\frac{k^n}{k!} \\
&=\sum_{k=1}^{\infty}\frac{k^n}{k!} \\
&= \sum_{k=1}^{\infty}\frac{((k-1)+1)^n}{k!}\\
&= \sum_{k=0}^{\infty}\frac{(k+1)^n}{(k+1)!} &k\to k+1\\
&= \sum_{k=0}^{\infty} \frac{(k+1)^{n-1}}{k!} &\text{simplifying}\\
 &=\sum_{k=0}^{\infty}\frac{1}{k!}\left( \sum_{j=0}^{n-1}\binom{n-1}{j}k^j\right) &\color{green}{\text{binomial expansion}}\\
&= \sum_{j=0}^{n-1}\binom{n-1}{j}\sum_{k=0}^{\infty}\frac{k^j}{k!} &\color{red}{\text{rearranging}} \\ 
&=  \sum_{j=0}^{n-1}\binom{n-1}{j} S_j
\end{aligned}
\end{align}
Now, if you are already familiar with Bell numbers, this will certainly ring a bell; otherwise it's a useful recursion relation to calculate the sum in the general case.
For $n=0$, by definition we know it's Euler's constant $S_0=e$. For $n=1$, we have:
$$S_1=\binom{0}{0}S_0=1 \times e = e$$
For $n=2$, which happens to be what has been asked, we have:
$$S_2=\binom 1 0 S_0 + \binom 1 1 S_1 = 2e$$
And so on. In general, since $S_n$'s have the same linear recursion relations as Bell numbers $B_n$, the solution would be $S_n=e B_n$. 
A beautiful way of constructing these Bell numbers would be using the triangles which I've found in this post:
\begin{array}{ccccccccccccc}
1 & & 1 & & 2 & & \color{red}5 & & 15 & & 52 & & \color{blue}{203} \\
& 2 & & 3 & & 7 & & 20 & & 67 & & 255 \\
& & \color{red}{5} & & 10 & & 27 & & 87 & & 322 \\
& & & 15 & & 37 & & 114 & & 409 \\
& & & & 52 & & 151 & & 523 \\
& & & & & \color{blue}{203} & & 674 \\
& & & & & & 877
\end{array}
A: No theory needed; you can solve this with simple manipulations.
$$
\sum_{1}^{\infty} \frac{k^2}{k!} = \sum_{1}^{\infty} \frac{k}{(k-1)!} = \sum_{0}^{\infty} \frac{k+1}{k!} = \sum_0^{\infty} \frac{1}{k!} + \sum_1^{\infty} \frac{k}{k!} = e + \sum_0^{\infty} \frac{1}{k!} = 2e
$$
A: $e^x=1+x+\dfrac{x^2}{2!}+\cdots+\dfrac{x^k}{k!}+\cdots$
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,
$2e=\dfrac{1^2}{1!}+\dfrac{2^2}{2!}+\cdots+\dfrac{k^2}{k!}+\cdots$ 
