# 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?

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

• I wonder...how efficiently can use this to solve for $$\sum_{n=0}^\infty\frac{n^k}{n!}$$ Mar 1, 2017 at 17:18
• @SimplyBeautifulArt it cannot. Look at the format of your series, $n$ is the summatory variable, but here the only thing you have control over in your desired series is $k$, unlike in the series for $e^x$ where you control the base rather than the exponent. Apr 11, 2017 at 14:49
• @AdamHughes I suppose you can get a recursive relationship for each value of $k$ though. Apr 11, 2017 at 15:29

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 $x=1$

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

• Perhaps you could see my more general answer, which takes the essence of your answer to arbitrary naturals. Mar 1, 2017 at 17:17
• @SimplyBeautifulArt I'm well aware of the method you use, but it was wildly inappropriate for the op's context (tagged as GRE question) so I opted for an approach that he would be most easily able to understand. Generalities are nice for oneself, but don't always help explain something to a student without the right background. Mar 1, 2017 at 17:22
• Oh, GRE... hm... I've actually never seen that tag :D Mar 1, 2017 at 17:28

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.