Considering the following summation of series: $$S_n=\sum_{k=0}^{n}(-1)^k{{n}\choose{k}}\sum_{m=0}^{k}(-1)^m\frac{k!}{(k-m)!}b^{-m},$$ where $n$ is a non-negative integer, and $b$ is a known non-zero constant.

I computed manually and got $$S_1=b^{-1}, S_2=2b^{-2}, S_3=6b^{-3}.$$ Then I set a hypothesis of $S_n$: $$S_n=n!\cdot b^{-n}.$$ However, I couldn't prove whether it is correct or not. Could someone help me? Thanks very much indeed!

  • 1
    $\begingroup$ We can reverse the order of summation and do some simplification to get $$ S_n=\sum_{m=0}^n\sum_{k=m}^n(-1)^k{n\choose k}(-1)^m{k!\over (k-m)!}b^{-m}=\sum_{m=0}^n(-1)^m m! b^{-m}\sum_{k=0}^n(-1)^k{n\choose k}{k\choose m}. $$ Hopefully this gives you a good idea on how to do it $\endgroup$ – munchhausen Jun 24 '18 at 9:24
  • $\begingroup$ @Munchhausen Thanks for your help :) $\endgroup$ – yuhou CHEN Jun 24 '18 at 10:31

Change the order of summation, then shift the summation index $k$ to $k-m$ and then use the binomial theorem:

\begin{align*}&\hskip-2cm\sum_{k=0}^{n}(-1)^k{{n}\choose{k}}\sum_{m=0}^{k}(-1)^m\frac{k!}{(k-m)!}b^{-m}\\&= \sum_{0\le m\le k \le n}(-1)^k \frac{n!}{k!(n-k)!}(-1)^m\frac{k!(n-m)!}{(k-m)!(n-m)!}b^{-m}\\ &= \sum_{m=0}^{n}(-1)^mb^{-m}\frac{n!}{(n-m)!}\sum_{k=m}^{n}(-1)^k\binom{n-m}{k-m}\\ &=\sum_{m=0}^{n}b^{-m}\frac{n!}{(n-m)!}\sum_{k=0}^{n-m}(-1)^{k}\binom{n-m}{k}\\ &=\sum_{m=0}^{n}b^{-m}\frac{n!}{(n-m)!}(1-1)^{n-m}\\ &=\sum_{m=0}^{n}b^{-m}\frac{n!}{(n-m)!}\delta_{n,m}\\ &=n!b^{-n}.\end{align*}

  • $\begingroup$ Thanks for your derivation! But I couldn't understand the first step well, How could the second expression come out after changing the order of summation? Could you give me more references? I'm not familiar with binomial theorem. And I have no question about the other steps. $\endgroup$ – yuhou CHEN Jun 24 '18 at 10:30
  • $\begingroup$ Could you please check my another question link? I'm wondering if that summation of series obtains the distributive law. Thanks a lot! $\endgroup$ – yuhou CHEN Jun 24 '18 at 13:26
  • $\begingroup$ @yuhouCHEN I have added an intermediate step. $\endgroup$ – Phira Jun 24 '18 at 14:42
  • $\begingroup$ That's excellent! Thank you so much! $\endgroup$ – yuhou CHEN Jun 24 '18 at 15:32

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