<s>[INST] Infinite Series $\sum\limits_{n=1}^\infty\left(\frac{H_n}n\right)^2$ How can I find a closed form for the following sum?
$$\sum_{n=1}^{\infty}\left(\frac{H_n}{n}\right)^2$$
($H_n=\sum_{k=1}^n\frac{1}{k}$).
 [/INST] This result was a problem in American mathematical monthly in 50s.
Source:
H. F. Sandham and Martin Kneser, The American mathematical monthly, Advanced problem 4305, Vol. 57, No. 4 (Apr., 1950), pp. 267-268
\begin{align}S&=\sum_{n=1}^\infty \left(\frac{\text{H}_n}{n}\right)^2\\
&=\sum_{n=1}^\infty \frac{1}{n^2}\left(\int_0^1 \frac{1-t^n}{1-t}dt\right)\left(\int_0^1 \frac{1-u^n}{1-u}du\right)\\
&=\int_0^1 \int_0^1 \frac{\text{Li}_2(1)+\text{Li}_2( tu)-\text{Li}_2(t)-\text{Li}_2(u)}{(1-t)(1-u)}dtdu\\
&\overset{\text{IBP}}=\int_0^1 \int_0^1 \frac{\ln(1-t)\big(\ln(1-t)-\ln(1-tu)\big)}{t(1-u)}dtdu\\ 
&\overset{x=1-tu,y=\frac{1-t}{1-tu}}=\int_0^1\int_0^1 \frac{\ln y\ln(xy)}{(1-y)(1-xy)}dxdy\\
&\overset{w(x)=yx}=\int_0^1 \frac{\ln y}{y(1-y)}\left(\int_0^y \frac{\ln w}{1-w}dw\right)dy\\
&=\int_0^1 \frac{\ln y}{1-y}\left(\int_0^y \frac{\ln w}{1-w}dw\right)dy+\underbrace{\int_0^1 \frac{\ln y}{y}\left(\int_0^y \frac{\ln w}{1-w}dw\right)dy}_{\text{IBP}}\\
&=\frac{1}{2}\left(\int_0^1 \frac{\ln w}{1-w}dw\right)^2-\frac{1}{2}\int_0^1 \frac{\ln^3 w}{1-w}dw\\
&=\frac{1}{2}\times \left(\frac{\pi^2}{6}\right)^2-\frac{1}{2}\times -\frac{\pi^4}{15}\\
&=\boxed{\dfrac{17\pi^4}{360}}.
\end{align} NB:
I assume that:
\begin{align}\zeta(2)=\frac{\pi^2}{6},&&\zeta(4)=\frac{\pi^4}{90},&&\int_0^1 \frac{\ln w}{1-w}dw=-\zeta(2)=-\frac{\pi^2}{6},&&\int_0^1 \frac{\ln^3 w}{1-w}dw=-6\zeta(4)=-\frac{\pi^4}{15}\end{align}
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