Compute $\sum\limits_{k=0}^n (-1)^k \binom{n}{k}\frac{1}{s+k}$ 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
$$
\sum_{k=0}^n (-1)^k \binom{n}{k}\frac{1}{s-k}=\dfrac{n!}{s(s-1)(s-2)...(s-n)}=\dfrac{\Gamma(n+1)\Gamma(s-n)}{\Gamma(s+1)}.
$$
 A: 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
$$
\sum_{k=0}^n (-1)^k \binom{n}{k}\frac{1}{s+k}=\int_0^1t^{s-1}(1-t)^ndt=\mathrm{Beta}(s,n+1)=\frac{\Gamma(s)\Gamma(n+1)}{\Gamma(s+n+1)}.$$
A: 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 $f(s) = g(s).$
A: prove by induction that
$\sum_{k=0}^{n}(-10^k\binom{n}{k}\frac{1}{s+k}={\frac {n!\,s!}{s \left( n+s \right) !}}$
