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?

An overkill. From the derivative version of the Melzak's identity $$\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}\frac{f\left(y-k\right)}{\left(x+k\right)^{2}}=\frac{f\left(x+y\right)\sum_{k=0}^{n}\frac{1}{k+x}-\frac{d}{dx}f\left(x+y\right)}{x\dbinom{x+n}{n}},\,x\neq-k$$ where $f$ is a polynomial up to degree $n$ we have, taking $f\left(z\right)\equiv1,\,x=1$, that $$\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}\frac{1}{\left(1+k\right)^{2}}=\frac{\sum_{k=0}^{n}\frac{1}{k+1}}{n+1}=\color{red}{\frac{H_{n+1}}{n+1}}$$ as wanted.
