Sum on integration and binomial theorem.

Question

If $(1+x)^n = \sum_{r=0}^n \binom{n}{r}x^r$ and $$\sum_{r=0}^n \frac{(-1)^r}{(r+1)^2} \binom{n}{r} = k\sum_{r=0}^n \frac{1}{r+1}$$

Then prove that $$k=\frac{1}{n+1}.$$

Answer 1

$$\begin{eqnarray*}\sum_{r=0}^{n}\frac{(-1)^r}{(r+1)^2}\binom{n}{r} &=& \iint_{(0,1)^2}\sum_{r=0}^{n}(-1)^r x^r y^r \binom{n}{r}\,dx\,dy \\&=&\iint_{(0,1)^2}(1-xy)^n\,dx\,dy\\&=&\frac{1}{n+1}\int_{0}^{1}\frac{1-(1-x)^{n+1}}{x}\,dx\\&=&\frac{1}{n+1}\int_{0}^{1}\frac{1-x^{n+1}}{1-x}\,dx\\&=&\frac{1}{n+1}\sum_{j=0}^{n}\int_{0}^{1}x^j\,dx\\&=&\frac{1}{n+1}\sum_{j=0}^{n}\frac{1}{j+1}\\&=&\frac{H_{n+1}}{n+1}.\end{eqnarray*}$$