Closed form for $\sum_{k=0}^{n} \binom{n}{k}\frac{(-1)^k}{(k+1)^2}$ How can I calculate the following sum involving binomial terms:
$$\sum_{k=0}^{n} \binom{n}{k}\frac{(-1)^k}{(k+1)^2}$$
Where the value of n can get very big (thus calculating the binomial coefficients is not feasible).
Is there a closed form for this summation?

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
$$\bbox[5px,border:2px solid #00A000]{
\frac{H_{n+1}}{n+1}.}$$
