I have to find the following sum, and I've solved up to a certain point but cannot proceed further. Any hints on how to proceed would be helpful.
$$\sum_{r=0}^{n}\frac{(-1)^r}{(r+1)^2}{n\choose r}$$
Consider the expansions of $(1-x)^{n}$ and $(1+x)^{n}$.
$$\begin{align}\int_{0}^{x}(1-x)^{n}\mathrm dx&=\sum_{r=0}^{n}\frac{(-1)^{r}x^r}{r+1}{n\choose r}\\\frac{1-(1-x)^{n+1}}{n+1}&=\sum_{r=0}^{n}\frac{(-1)^{r}x^r}{r+1}{n\choose r}\tag1\end{align}$$
Similarly integrating $(1+x)^{n}$ from $0$ to $x$.
$$\frac{(1+x)^{n+1}-1}{n+1}=\sum_{r=0}^{n}\frac{x^r}{r+1}{n\choose r}\tag2$$
Replacing $x\mapsto 1/x$ in $(2)$, the required summation is just the coefficient of $x^{0}$ in the product of $(1)$ and $(2)$, which simplifies to the following.
$$\text{coeff. of }x^{n+1} \text{ in } \left[\underbrace{(1+x^{n+1}-x^{n+1})}_{=0}-(1-x^2)^{n+1}+\underbrace{x^{n+1}(1-x)^{n+1}}_{=1}\right]$$
Any hints on how to proceed further. Thanks.
