Compute $\sum\limits_{k=0}^n (-1)^k \binom{n}{k}\frac{1}{s+k}$

Question

Anyone can help me finding this summation:

$$ \sum_{k=0}^n (-1)^k \binom{n}{k}\frac{1}{s+k}. $$ Where there is a similar one with known answer $$ \sum_{k=0}^n (-1)^k \binom{n}{k}\frac{1}{s-k}=\dfrac{n!}{s(s-1)(s-2)...(s-n)}=\dfrac{\Gamma(n+1)\Gamma(s-n)}{\Gamma(s+1)}. $$

Answer 1

Note that, for every positive $s$, $$ \sum_{k=0}^n (-1)^k \binom{n}{k}\frac{1}{s+k}= \sum_{k=0}^n (-1)^k \binom{n}{k}\int_0^1t^{s+k-1}dt=\int_0^1t^{s-1}\sum_{k=0}^n (-1)^k \binom{n}{k}t^kdt$$ hence $$ \sum_{k=0}^n (-1)^k \binom{n}{k}\frac{1}{s+k}=\int_0^1t^{s-1}(1-t)^ndt=\mathrm{Beta}(s,n+1)=\frac{\Gamma(s)\Gamma(n+1)}{\Gamma(s+n+1)}.$$

Answer 2

prove by induction that $\sum_{k=0}^{n}(-10^k\binom{n}{k}\frac{1}{s+k}={\frac {n!\,s!}{s \left( n+s \right) !}}$

Answer 3

Introduce the rational function $f(s)$ with poles at $s=0,-1,-2,\ldots -n:$ $$f(s) = \sum_{k=0}^n (-1)^k {n\choose k} \frac{1}{s+k}.$$ Now observe that $$\mathrm{Res}_{s=-k} f(s) = (-1)^k {n\choose k}.$$

Compare with $g(s)$ given by $$g(s) = \frac{n!}{s(s+1)(s+2)\cdots(s+n)}.$$ The poles of $g(s)$ are the same as the poles of $f(s)$ and both are simple, with residue $$\mathrm{Res}_{s=-k} g(s) = \frac{n!}{(-k)(-k+1)\cdots(-k+k-1)(-k+k+1)(-k+k+2)\cdots(-k+n)}.$$ This simplifies to $$\mathrm{Res}_{s=-k} g(s) = \frac{n! (-1)^k}{k!(n-k)!} = (-1)^k {n\choose k}.$$

We conclude that $f(s) = g(s).$