Series representation of $2e$ According to GR9768, problem 37: $$\sum_{k=1}^{+\infty} \frac{k^2}{k!} = 2e$$
Can someone please explain how to get started in showing that?
 A: $$\sum \limits_{k = 1}^\infty \frac{k^2}{k!} = \sum \limits_{k = 1}^\infty \frac{k}{(k - 1)!} = \sum \limits_{k = 0}^\infty \frac{k + 1}{k!} = \sum \limits_{k = 1}^\infty \frac{1}{(k - 1)!} + \sum \limits_{k = 0}^\infty \frac{1}{k!} = 2 \sum \limits_{k = 0}^\infty \frac{1}{k!} = 2e$$
A: Write the series
$$e^{x}=\sum_{k=0}^\infty {x^k\over k!}$$
Now apply the derivative and multiply by $x$ to get
$$x{d\over dx}(e^x) = xe^x = \sum_{k=0}^\infty {kx^k\over k!}$$
Now do it again
$$x{d\over dx} (xe^x) = e^x + xe^x = \sum_{k=0}^\infty {k^2x^k\over k!}$$
Now let $x=1$

$$e^1+1\cdot e^1 =2e = \sum_{k=0}^\infty {k^2\over k!}$$

A: Consider the following:
$$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$$
Let $x\mapsto e^x$:
$$e^{e^x}=\sum_{n=0}^\infty\frac{e^{nx}}{n!}$$
It thus follows that
$$\frac{d^k}{dx^k}e^{e^x}\bigg|_{x=0}=\sum_{n=0}^\infty\frac{n^k}{n!}$$
And by applying Faà di Bruno's formula, we find that
$$eB_k=\sum_{n=0}^\infty\frac{n^k}{n!}$$
Where $B_k$ is the $k$th Bell number.
This is famously known as Dobiński's formula.
