Suppose we seek to compute
$S_n = \sum_{k=0}^n {n\choose k} \frac{(-1)^k}{(k+1)^2}.$
With this in mind we introduce the function
$f(z) = n! (-1)^n \frac{1}{(z+1)^2} \prod_{q=0}^n \frac{1}{z-q}.$
We then obtain for $0\le k\le n$
$\mathrm{Res}_{z=k} f(z) = (-1)^n \frac{n!}{(k+1)^2} \prod_{q=0}^{k-1} \frac{1}{k-q} \prod_{q=k+1}^n \frac{1}{k-q} \\ = (-1)^n \frac{n!}{(k+1)^2} \frac{1}{k!} \frac{(-1)^{n-k}}{(n-k)!} = {n\choose k} \frac{(-1)^k}{(k+1)^2}.$
This means that
$S_n = \sum_{k=0}^n \mathrm{Res}_{z=k} f(z)$
and since residues sum to zero we have
$S_n + \mathrm{Res}_{z=-1} f(z) + \mathrm{Res}_{z=\infty} f(z) = 0.$
We can compute the residue at infinity by inspection (it is zero) or more formally through
$\mathrm{Res}_{z=\infty} n! (-1)^n \frac{1}{(z+1)^2} \prod_{q=0}^n \frac{1}{z-q} \\ = - n! (-1)^n \mathrm{Res}_{z=0} \frac{1}{z^2} \frac{1}{(1/z+1)^2} \prod_{q=0}^n \frac{1}{1/z-q} \\ = - n! (-1)^n \mathrm{Res}_{z=0} \frac{1}{(z+1)^2} \prod_{q=0}^n \frac{z}{1-qz} \\ = - n! (-1)^n \mathrm{Res}_{z=0} \frac{z^{n+1}}{(z+1)^2} \prod_{q=0}^n \frac{1}{1-qz} = 0.$
We get for the residue at $z=-1$ that
$\mathrm{Res}_{z=-1} f(z) = n! (-1)^n \left. \left(\prod_{q=0}^n \frac{1}{z-q}\right)'\right|_{z=-1} \\ = - n! (-1)^n \left. \left(\prod_{q=0}^n \frac{1}{z-q}\right) \sum_{q=0}^n \frac{1}{z-q} \right|_{z=-1} \\ = - n! (-1)^n \frac{(-1)^{n+1}}{(n+1)!} \left(-H_{n+1}\right) = -\frac{H_{n+1}}{n+1}.$
We thus have
$S_n -\frac{H_{n+1}}{n+1} = 0$
or
$\bbox[5px,border:2px solid #00A000]{ \frac{H_{n+1}}{n+1}.}$