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?


3 Answers 3


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

  • $\begingroup$ I wonder...how efficiently can use this to solve for $$\sum_{n=0}^\infty\frac{n^k}{n!}$$ $\endgroup$ Mar 1, 2017 at 17:18
  • $\begingroup$ @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. $\endgroup$ Apr 11, 2017 at 14:49
  • $\begingroup$ @AdamHughes I suppose you can get a recursive relationship for each value of $k$ though. $\endgroup$ 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 $x=1$

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

  • $\begingroup$ Perhaps you could see my more general answer, which takes the essence of your answer to arbitrary naturals. $\endgroup$ Mar 1, 2017 at 17:17
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    $\begingroup$ @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. $\endgroup$ Mar 1, 2017 at 17:22
  • $\begingroup$ Oh, GRE... hm... I've actually never seen that tag :D $\endgroup$ Mar 1, 2017 at 17:28

Consider the following:


Let $x\mapsto e^x$:


It thus follows that


And by applying Faà di Bruno's formula, we find that


Where $B_k$ is the $k$th Bell number.

This is famously known as Dobiński's formula.


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