# Finite summation of series involving binomial coefficients

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!

• 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 – munchhausen Jun 24 '18 at 9:24
• @Munchhausen Thanks for your help :) – 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: