TITLE: Show that : $\sum_{k=0}^{n}{(-1)^k{n\choose k}\over (k+1)^2}={1+\sqrt{1+4(n+1)^2H_nH_{n+1}}\over 2(n+1)^2}$, for $n\ge0$
QUESTION [5 upvotes]: Show that
$$\sum_{k=0}^{n}{(-1)^k{n\choose k}\over (k+1)^2}={1+\sqrt{1+4(n+1)^2H_nH_{n+1}}\over 2(n+1)^2}$$
$H_n$; is the n-th harmonic number
My try:
We know that
$$\sum_{k=0}^{\infty}{(-1)^k\over (k+1)^2}=\int_{0}^{1}{-\ln{x}\over 1+x}dx$$
But
$$\sum_{k=0}^{n}{(-1)^k{n\choose k}\over (k+1)^2}=\int_{0}^{a}???$$
From wikipedia I found 
$$H_n=\int_{0}^{1}{1-x^n\over 1-x}dx=\sum_{k=1}^{n}(-1)^{k-1}{{n\choose k}\over k}$$ may be of useful
Onward I am clueless as what to do next. Please help!

REPLY [3 votes]: Since $\frac{1}{(1+k)^2}=\int_{0}^{1} x^k(-\log x)\,dx $ we have:
$$\begin{eqnarray*} \sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^k}{(k+1)^2} &=& \int_{0}^{1}(-\log x)\sum_{k=0}^{n}\binom{n}{k}(-x)^k\,dx\\
&=&\int_{0}^{1}(-\log x)(1-x)^n\,dx\\&=&-\int_{0}^{1}x^n\log(1-x)\,dx\\&=&-\left.\frac{d}{d\alpha}\int_{0}^{1}x^n(1-x)^{\alpha}\,dx\,\right|_{\alpha=0^+}\\&=&-\left.\frac{d}{d\alpha}\frac{\Gamma(n+1)\,\Gamma(\alpha+1)}{\Gamma(n+\alpha+1)}\,\right|_{\alpha=0^+}\\&=&\frac{H_{n+1}}{n+1} \end{eqnarray*}$$
and since $H_{n+1}=H_n+\frac{1}{n+1}$,
$$ 1+4(n+1)^2 H_n H_{n+1} = 1+4(n+1) H_n + 4(n+1)^2 H_n^2 = (1+2(n+1)H_n)^2.$$