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?

$(1-x)^n=\sum_{0\le k\le n}\binom nk(-1)^kx^k$
Integrating wrt $x,$ 
$$-\frac{(1-x)^{n+1}}{n+1}+C=\sum_{0\le k\le n}\binom nk(-1)^k\frac{x^{k+1}}{k+1}$$ where $C$ is the indefinite constant.
Putting $x=0,C-\frac1{n+1}=0\implies C=\frac1{n+1}$
So, $$\sum_{0\le k\le n}\binom nk(-1)^k\frac{x^{k+1}}{k+1}=\frac1{n+1}-\frac{(1-x)^{n+1}}{n+1}$$
So, $$\sum_{0\le k\le n}\binom nk(-1)^k\frac{x^{k}}{k+1}=\frac1{x(n+1)}-\frac{(1-x)^{n+1}}{(n+1)x}$$
Again integrating wrt $x,$
$$\sum_{0\le k\le n}\binom nk(-1)^k\frac{x^{k+1}}{(k+1)^2}=\frac {\log x}{n+1}-\int\frac{(1-x)^{n+1}}{(n+1)x}dx+D$$ where $D$ is the indefinite constant.
