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

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 $$ \sum_{k=0}^n (-1)^k \binom{n}{k}\frac{1}{s+k}=\int_0^1t^{s-1}(1-t)^ndt=\mathrm{Beta}(s,n+1)=\frac{\Gamma(s)\Gamma(n+1)}{\Gamma(s+n+1)}.$$