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This question already has an answer here:

So, I wonder what is the evaluation of $$\sum_{k = 0}^{n} {n\choose k} k^m\text{,}\qquad (*)$$ where $m,n\in \mathbb{N}$.

One of my tries: knowing that $$k^m = \sum_{j = 0}^{m}\text{S}(m,j)\cdot k(k-1)\cdots(k-j+1)\text{,}$$ where $\text{S}(m,k)$ are Stirling numbers of the second kind, and $$\sum_{k = 0}^n {n\choose k} k(k-1)\cdots(k-j+1) =2^{n-j}\cdot n(n-1)\cdots(n-j+1)\text{,}$$ I have rewritten the upper sum into $$\sum_{k = 0}^{n}\sum_{j = 0}^m {n\choose k} \text{S}(m,j)\cdot k(k-1)\cdots(k-j+1)\text{.}$$ Changing the order of summation, we get $$\sum_{j = 0}^{m}\text{S}(m,j) \cdot 2^{n-j}\cdot n(n-1)\cdots (n-j+1)\text{,}$$ and here it stops.

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marked as duplicate by Sil, Lee David Chung Lin, Alexander Gruber Nov 5 '18 at 3:26

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.

  • 1
    $\begingroup$ Since this was never mentioned, I'll throw it out there: there is almost certainly no closed-form description of this sum, because this would give a closed form generating function for the Stirling numbers. For any given $m$, what you've written is a closed form, and this is the best we can hope for. $\endgroup$ – Slade Sep 8 '14 at 2:57
  • $\begingroup$ This sum was discussed at this MSE link. $\endgroup$ – Marko Riedel Jul 2 '15 at 20:20
  • $\begingroup$ For more than two years, we were not aware of the duplication. Wow. $\endgroup$ – Antoine Nov 10 '18 at 19:06
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related techniques: (I), (II). Here is how you advance. Recalling the identities

$$ \sum_{k = 0}^{n} {n\choose k} x^k = (1+x)^n, $$

and

$$ (xD)^m = \sum_{s=0}^{m} {m\brace s} x^s D^s, $$

where $D=\frac{d}{dx}$ and ${m\brace s}$ are Stirling numbers of the second kind. Now, apply the above operator to both sides of the first equation as

$$ (xD)^m \sum_{k = 0}^{n} {n\choose k} x^k = \sum_{s=0}^{m} {m\brace s} x^s D^s (1+x)^n. $$

I leave it for you to finish the problem.

Note:

$$ D^s (1+x)^n = \frac{\Gamma(n+1)}{\Gamma(n-s+1)}(1+x)^{n-s} .$$

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  • $\begingroup$ I apply the operator $(xD)^m$ (by definition) to the left side of the first equality and to it's right side (using the second identity). All I get is $$\sum_{s = 0}^m {n\choose k}k^m x^k = \sum_{s = 0}^{m}{m\brace s}x^s(1+x)^{n-s}n^{\underline{s}}\text{,}$$ where $n^\underline{s} = n(n-1)\cdots (n-s+1)$. For $x=1$, this is the equality from the question. What am I missing? $\endgroup$ – Antoine Mar 16 '14 at 14:53
  • $\begingroup$ @Antoine: You need to substitute $x=1$ to get your series. $\endgroup$ – Mhenni Benghorbal Mar 16 '14 at 15:09
  • $\begingroup$ Yes, I know that. Maybe I should point out that I am aware of equality $$\sum_{s = 0}^m {n\choose k}k^m = \sum_{s = 0}^{m}{m\brace s}\cdot 2^{n-s}n^{\underline{s}}\text{,}$$ (I think this is evident from the formulation of my question). I have also noticed that substitution $x= 1$ gives these two equalities (I think this is evident from my previous comment). I just want the evaluation/close form of one of these sums. $\endgroup$ – Antoine Mar 16 '14 at 15:39
  • $\begingroup$ Ok, maybe my awareness of equality was not evident from my question. I have edited it. $\endgroup$ – Antoine Mar 16 '14 at 15:55

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