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This question already has an answer here:

I stumbled upon a similar problem and really liked the answers there, so I wondered if there were a general solution for

$$\sum_{k=1}^{\infty}\frac{k^n}{k!}=?$$

Sadly, when I try to apply some of the methods from the answers on the other problem, I fall short of a complete answer.

My attempt:

$$e^x=\sum_{k=1}^{\infty}\frac{x^k}{k!}$$

Differentiate once, then multiply both sides by $x$.

$$xe^x=\sum_{k=1}^{\infty}\frac{kx^k}{k!}$$

Differentiate both sides again, then multiply by $x$.

$$x(x+1)e^x=\sum_{k=1}^{\infty}\frac{k^2x^k}{k!}$$

Repeat this process $n$ times to find the general solution for $\sum_{k=1}^{\infty}\frac{k^n}{k!}$ (substituting $x=1$), however, I cannot see a way to proceed like this.

UPDATE

I've found that Dobinski's formula is directly related, and that $\sum_{k=1}^{\infty}\frac{k^n}{k!}=eB_n$

Where $B_n$ is Bell's numbers.

I'll leave this question open for anyone who wants to attempt the problem using derivatives.

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marked as duplicate by J. M. is a poor mathematician, John B, user99914, Jack's wasted life, user91500 Mar 25 '16 at 3:33

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ When you write $k^n$, what is $n$? $\endgroup$ – Arthur Mar 24 '16 at 23:57
  • $\begingroup$ @Arthur $n\in\mathbb{N}$ $\endgroup$ – Simply Beautiful Art Mar 24 '16 at 23:57
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    $\begingroup$ Just to be clear: $e^x=\sum_{k=0}^{\infty} \frac {x^k}{k!}$. Your sums start at $k=1$. $\endgroup$ – lulu Mar 25 '16 at 0:00
  • $\begingroup$ @lulu Actually, it doesn't matter since $\sum_{k=0}^{\infty}\frac{k^n}{k!}=\sum_{k=1}^{\infty}\frac{k^n}{k!}$ $\endgroup$ – Simply Beautiful Art Mar 25 '16 at 0:00
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    $\begingroup$ This is about as poorly motivated as you can get but, with playing around with Mathematica, I think that $\sum_{k=1}^\infty \frac{k^n}{k!} = e B_n$, where $B_n$ is the $n$'th Bell number. Wikipedia has an article on them. $\endgroup$ – fred Mar 25 '16 at 0:03
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If you compute the series with the very first terms you'll get for the series:

$$S = \sum_{k = 1}^{+\infty}\frac{k^n}{k!}$$

$n = 1 ~~~~~ \to ~~~~~ S = e$

$n = 2 ~~~~~ \to ~~~~~ S = 2e$

$n = 3 ~~~~~ \to ~~~~~ S = 5e$

$n = 4 ~~~~~ \to ~~~~~ S = 15e$

$n = 5 ~~~~~ \to ~~~~~ S = 52e$

$n = 6 ~~~~~ \to ~~~~~ S = 203e$

$n = 7 ~~~~~ \to ~~~~~ S = 877e$

$n = 8 ~~~~~ \to ~~~~~ S = 4140e$

Looking at the coefficient in front of "e" in every term, we recognize the famous succession

$$1, 2, 5, 15, 52, 203, 877, 4140, 21147, 115975, 678570, \cdots $$

Which are the famous Bell's Numbers. Then your series can be compute in this way:

$$S = \sum_{k = 1}^{+\infty}\frac{k^n}{k!} = \mathcal{B}_n\ e$$

Where $\mathcal{B}_n$ is the $n$-th Bell's number. This clearly depends on what $n$ you take initially in the series, then it's easily evaluable.

More here

https://en.wikipedia.org/wiki/Bell_number

http://mathworld.wolfram.com/BellNumber.html

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  • $\begingroup$ Yes... I found this to be the case, as you may have noticed. In particular, this is Dobinski's formula... I really just wanted a different result, as I have already made this realization. $\endgroup$ – Simply Beautiful Art Mar 25 '16 at 0:56
  • $\begingroup$ Whoops, I never read comments below questions, sorry :D Anyway I don't think there does exist another close result. Bell's numbers are the most exact result one can get. For what I know.. $\endgroup$ – Von Neumann Mar 25 '16 at 0:59
  • $\begingroup$ Thanks! I'll just leave this question open for future answer-ers. $\endgroup$ – Simply Beautiful Art Mar 25 '16 at 1:00

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