<s>[INST] Evaluate $1+\left(\frac{1+\frac12}{2}\right)^2+\left(\frac{1+\frac12+\frac13}{3}\right)^2+\left(\frac{1+\frac12+\frac13+\frac14}{4}\right)^2+...$ Evaluate:
$$S_n=1+\left(\frac{1+\frac12}{2}\right)^2+\left(\frac{1+\frac12+\frac13}{3}\right)^2+\left(\frac{1+\frac12+\frac13+\frac14}{4}\right)^2+...$$
a_n are the individual terms to be summed.
My Try :
\begin{align}
&a_1=1\\
&a_2=\left(\frac{3}{4}\right)^2=\frac{9}{16}\\
&a_3=\left(\frac{11}{18}\right)^2\\
&a_4=\left(\frac{25}{48}\right)^2
\end{align}
now :?
 [/INST] By setting $H_n = \sum_{k=1}^{n}\frac{1}{k}$ we have to compute $\sum_{n\geq 1}\left(\frac{H_n}{n}\right)^2$. We may notice that
$$ \sum_{n=1}^{N}\frac{H_n}{n}=\sum_{1\leq m\leq n\leq N}\frac{1}{mn}=\frac{H_N^2+H_N^{(2)}}{2}\tag{1}$$
and for the same reason
$$ \sum_{n=1}^{N}\frac{H_n^{(2)}}{n^2} = \frac{1}{2}\left[\left(\sum_{n=1}^{N}\frac{1}{n^2}\right)^2+\sum_{n=1}^{N}\frac{1}{n^4}\right]\stackrel{N\to +\infty}{\longrightarrow}\frac{\zeta(2)^2+\zeta(4)}{2}=\frac{7\pi^4}{360} \tag{2}$$
Since $-\log(1-x)=\sum_{n\geq 1}\frac{1}{n}\,x^n$, by multiplying both sides by $\frac{1}{1-x}$ and applying termwise integration we have
$$ \sum_{n\geq 1}\frac{H_{n}}{n}\,x^{n} = \text{Li}_2(x)+\frac{1}{2}\log^2(1-x) \tag{3}$$
hence by $(1)$ it follows that:
$$ \sum_{N\geq 1}\frac{H_N^2+H_N^{(2)}}{2}x^{N} = \frac{\text{Li}_2(x)}{1-x}+\frac{1}{2}\cdot\frac{\log^2(1-x)}{1-x}\tag{4} $$
and by multiplying both sides of $(4)$ by $-\frac{2\log x}{x}$ and performing termwise integration over $(0,1)$:
$$ \sum_{N\geq 1}\frac{H_N^2+H_N^{(2)}}{N^2} = -\int_{0}^{1}\left[\frac{2\text{Li}_2(x)\log(x)}{x(1-x)}+\frac{\log^2(1-x)\log(x)}{x(1-x)}\right]\,dx.\tag{5} $$
The integral $-\int_{0}^{1}\frac{\log^2(1-x)\log(x)}{x(1-x)}\,dx$ can be computed by differentiating Euler's beta function, and it equals $\frac{\pi^4}{36}$. Since $\int\frac{\log(x)}{x(1-x)}\,dx=\frac{1}{2}\log^2(x)+\text{Li}_2(1-x)$ and $\frac{d}{dx}\text{Li}_2(x)=-\frac{\log(1-x)}{x}$, by integration by parts the whole problem boils down to computing:
$$ I = \int_{0}^{1}\frac{\text{Li}_2(x)\log(x)}{1-x}\,dx \tag{6}$$
but we have already done that in $(2)$, since $\frac{\text{Li}_2(x)}{1-x}=\sum_{n\geq 1}H_n^{(2)}x^n.$ Collecting pieces,

$$ \sum_{n\geq 1}\left(\frac{H_n}{n}\right)^2 = \color{red}{\frac{17\pi^4}{360}}.$$

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