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Proof that $$\sum_{k=1}^n (-1)^kk^m{n \choose k} = 0$$ for every $n>m$ and $m>1$.

I tried to use properties of ${n \choose k}.\;$ I tried to use $${n \choose k} = {n-1 \choose k} + {n-1 \choose k-1}$$ and using induction got stuck here $$\sum_{k=1}^t (-1)^kk^m{t-1 \choose k-1} = 0$$, where $t = n - 1$.

amWhy
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jack jones
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    Please show the work resulting from your try, and include it in your post. Show us the properties you used, how far you got in your try, and what specifically leaves you stuck after your claimed "try". – amWhy Nov 11 '17 at 15:00
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    You should probably also specify the values of $m$ that you're considering. I'm guessing that you mean for nonnegative or positive integers $m$. – RideTheWavelet Nov 11 '17 at 15:01
  • Thanks, jack jones, for the edit! – amWhy Nov 11 '17 at 15:09

3 Answers3

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This looks false. If $0=m<n=1$ we get $$\sum_{k=1}^n(-1)^kk^m\binom{n}{k} = -1^0\binom{1}{1}=-1\neq0\ .$$

Daniel Robert-Nicoud
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    check edited version – jack jones Nov 11 '17 at 15:09
  • And it might help to show what happens when $1\lt m\lt n$. – amWhy Nov 11 '17 at 15:10
  • Daniel The fact that you jumped into answer prior to knowing how m relates to n, (and not taking the time to even ask, (e.g., see the inequality above), lead you to posting a trivial and non-applicable argument. – amWhy Nov 11 '17 at 15:15
  • And why would you have even guessed that m could be $0$? – amWhy Nov 11 '17 at 15:18
  • @amWhy This is not my fault, but rather the OP's, since he didn't put all of the conditions in the question. – Daniel Robert-Nicoud Nov 11 '17 at 15:33
  • Daniel It is partly your fault for assuming things that were not stated in the question. You came up with a trivial counterexample, exploiting what wasn't said, and avoiding addressing the reasonable problem, and not having waited for further clarification. In any case, you jumped the gun, made assumptions that $m$ could be zero, to come up with what turns out to be an invalid claim that the statement is false, therefore. – amWhy Nov 11 '17 at 15:42
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Here are two hints that lead in different directions. First, consider the number of surjective functions from an $m$-element set to an $n$-element set. Second consider the coefficient of $x^my^n$ in the generating function $\exp(y(e^x - 1))$.

Sera Gunn
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We can show the identity with the help of generating functions. It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series. This way we can write for instance \begin{align*} [z^m]e^{kz}=[z^m]\left(1+kz+\frac{(kz)^2}{2!}+\cdots\right)=\frac{k^m}{m!}\tag{1} \end{align*}

We obtain for integers $1\leq m< n$ \begin{align*} \color{blue}{\sum_{k=0}^n\binom{n}{k}(-1)^kk^m}&=\sum_{k=0}^n\binom{n}{k}(-1)^{k}m![z^m]e^{kz}\tag{2}\\ &=(-1)^nm![z^m]\sum_{k=0}^n\binom{n}{k}\left(e^z\right)^k(-1)^{n-k}\tag{3}\\ &=(-1)^nm![z^m]\left(e^z-1\right)^n\tag{4}\\ &=(-1)^nm![z^m]\left(z+\frac{z^2}{2!}+\cdots\right)^n\tag{5}\\ &\color{blue}{=0 }\end{align*} and the claim follows.

Comment:

  • In (2) we use the coefficient of operator as shown in (1).

  • In (3) we do a small rearrangement.

  • In (4) we apply the binomial theorem.

  • In (5) we see the smallest power of $z$ is $n$.

Markus Scheuer
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