3

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.

Community
  • 1
Simply Beautiful Art
  • 76,603

1 Answers1

1

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

  • 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. – Simply Beautiful Art Mar 25 '16 at 00:56
  • 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.. –  Mar 25 '16 at 00:59
  • Thanks! I'll just leave this question open for future answer-ers. – Simply Beautiful Art Mar 25 '16 at 01:00