Series representation of $2e$

Question

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?

Answer 1

$$\sum \limits_{k = 1}^\infty \frac{k^2}{k!} = \sum \limits_{k = 1}^\infty \frac{k}{(k - 1)!} = \sum \limits_{k = 0}^\infty \frac{k + 1}{k!} = \sum \limits_{k = 1}^\infty \frac{1}{(k - 1)!} + \sum \limits_{k = 0}^\infty \frac{1}{k!} = 2 \sum \limits_{k = 0}^\infty \frac{1}{k!} = 2e$$

Answer 2

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 $x=1$

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

Answer 3

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.