Closed form for $ \frac{H_k}{k^2} $. I was trying to solve some problem and came across the following series:
$$
\sum_{k=2}^{\infty}\frac{H_k}{k^2}
$$
I tried to find a closed form for that series but could not. Also I looked some articles related to harmonic sum but it did not help me too.
Any hints and suggestions would be appreciated.
Thanks!
 A: Following the suggestion I gave above:
$$\begin{aligned}
\sum_{k=1}^{\infty}\frac{\mathcal{H}_k}{k^2} &=\sum_{k=1}^{\infty}\frac{1}{k^2}\int_{0}^{1}\frac{1-x^k}{1-x}\,dx \\ 
 &=\int_{0}^{1}\frac{1}{1-x} \sum_{k=1}^{\infty}\frac{1-x^k}{k^2}\,dx\\ 
 &= \int_{0}^{1}\frac{1}{1-x}\sum_{k=1}^{\infty}\left ( \frac{1}{k^2}-\frac{x^k}{k^2} \right )\,dx\\ 
 &= \int_{0}^{1}\frac{1}{1-x}\left ( \frac{\pi^2}{6}-\sum_{k=1}^{\infty}\frac{x^k}{k^2} \right )\,dx\\ 
 &=\int_{0}^{1}\frac{\frac{\pi^2}{6}-{\rm Li_2}(x)}{1-x}\,dx \\
 &=\ln(1-x)\left ( \frac{\pi^2}{6}-{\rm Li_2}(x) \right )\bigg|_0^1 + \int_{0}^{1}\frac{\ln^2 (1-x)}{x}\,dx \\
 &=\int_{0}^{1}\frac{\ln^2(1-x)}{x}\,dx \\
 &=2\zeta(3)
\end{aligned}$$
hence the original series evaluates to $2\zeta(3)-1$. 
The last integral is evaluated using the Taylor expansion of $\ln(1-x)$.
A: Asymptotic result is really easy: just notice $H_k \sim \log k$, so the resulting sum is a monotone decreasing function, hence it's asymptotically bounded by the corresponding integral: $\int_{1}^n \frac{\log x dx}{x^2}$ which is easily solved by IBP and the result is $O(\frac{\log n }{n^3})$.  
