Different approaches to solve nonlinear Euler sum

Question

I am looking for different approaches to prove the following series

$$\sum_{k=1}^\infty \frac{(H_k)^2}{k^2} = \frac{17\pi^4}{360}$$

My approach is using stirling numbers illustrate here.

What other approaches that could be used ?

Answer 1

Recall that the Dilogarithm function denoted by $\mathrm{Li}_2\left ( z \right )$, for $\left | z \right |\leq 1$ we have $$\mathrm{Li}_2\left ( z \right )=\sum_{n=1}^{\infty }\frac{z^{n}}{n^{2}}=-\int_{0}^{z}\frac{\ln\left ( 1-t \right )}{t}\, \mathrm{d}t$$ When $z = 1$ we have that $$\mathrm{Li}_2\left ( 1 \right )=\sum_{n=1}^{\infty }\frac{1}{n^{2}}=\frac{\pi ^{2}}{6}$$ If $x\in \left ( 0,1 \right )$, the following identity holds $$\ln x\ln\left ( 1-x \right )-\frac{1}{2}\ln^2\left ( 1-x \right )+\mathrm{Li}_2\left ( x \right )+\int_{1}^{1/\left ( 1-x \right )}\frac{\ln\left ( u-1 \right )}{u}\, \mathrm{d}u\tag1$$ To prove this identity we let $f:\left ( 0,1 \right )\rightarrow \mathbb{R}$ be the function defined by the left hand side of $(1)$. Then, a straightforward calculation shows that $f'(x)=0$. Thus, $f$ is a constant function and hence $\displaystyle f\left ( x \right )=\lim_{x\rightarrow 0}f\left ( x \right )=0$.

We also need the following integral as Zaid Alyafeai mentioned $$\int^1_0 \frac{\ln^3(1-x)}{x}\,\mathrm{d}x = -\frac{\pi ^{4}}{15}$$ Now we are ready to solve the problem. Based on $$\int_{0}^{1}x^{n}\ln\left ( 1-x \right )\mathrm{d}x=-\frac{H_{n+1}}{n+1}$$ we have \begin{align*} \sum_{n=1}^{\infty }\left ( \frac{H_n}{n} \right )^{2}&=\sum_{n=1}^{\infty }\int_{0}^{1}x^{n-1}\ln\left ( 1-x \right )\mathrm{d}x\int_{0}^{1}y^{n-1}\ln(1-y)\, \mathrm{d}y\\ &=\int_{0}^{1}\int_{0}^{1}\left ( \sum_{n=1}^{\infty }\left ( xy \right )^{n-1} \right )\ln\left ( 1-x \right )\ln\left ( 1-y \right )\mathrm{d}x\mathrm{d}y\\ &=\int_{0}^{1}\int_{0}^{1}\frac{\ln\left ( 1-x \right )\ln\left ( 1-y \right )}{1-xy}\mathrm{d}x\mathrm{d}y\\ &=\int_{0}^{1}\ln\left ( 1-x \right )\left (\int_{0}^{1}\frac{\ln\left ( 1-y \right )}{1-xy}\mathrm{d}y \right )\mathrm{d}x \end{align*} We calculate the inner integral, by making the substitution $1−xy = t$, and we have \begin{align*} \int_{0}^{1}\frac{\ln\left ( 1-y \right )}{1-xy}\mathrm{d}y&=\frac{1}{x}\int_{1-x}^{1}\frac{\ln\left ( 1-\dfrac{1}{x}+\dfrac{t}{x} \right )}{t}\, \mathrm{d}t\\ &=\frac{1}{x}\int_{1-x}^{1}\frac{\ln\left ( \dfrac{1}{x} \right )}{t}\, \mathrm{d}t+\frac{1}{x}\int_{1-x}^{1}\frac{\ln\left ( x-1+t \right )}{t}\, \mathrm{d}t\\ &=\frac{1}{x}\ln x\ln\left ( 1-x \right )+\frac{1}{x}\int_{1-x}^{1}\frac{\ln\left ( x-1+t \right )}{t}\, \mathrm{d}t \end{align*} The substitution $t =(1−x)u$, in the preceding integral, combined with $(1)$ implies that \begin{align*} \int_{0}^{1}\frac{\ln\left ( 1-y \right )}{1-xy}\mathrm{d}y&=\frac{1}{x}\ln x\ln\left ( 1-x \right )+\frac{1}{x}\int_{1}^{1/\left ( 1-x \right )}\frac{\ln\left ( 1-x \right )\left ( u-1 \right )}{u}\, \mathrm{d}u\\ &=\frac{1}{x}\left ( \ln x\ln\left ( 1-x \right )-\ln^2\left ( 1-x \right )+\int_{1}^{1/\left ( 1-x \right )}\frac{\ln\left ( u-1 \right )}{u}\, \mathrm{d}u \right )\\ &=\frac{1}{x}\left ( -\frac{1}{2}\ln^2\left ( 1-x \right )-\mathrm{Li}_2\left ( x \right ) \right ) \end{align*} It follows that \begin{align*} \sum_{n=1}^{\infty }\left ( \frac{H_n}{n} \right )^{2}&=-\frac{1}{2}\int_{0}^{1}\frac{\ln^3\left ( 1-x \right )}{x}\, \mathrm{d}x-\int_{0}^{1}\frac{\ln\left ( 1-x \right )\mathrm{Li}_2\left ( x \right )}{x}\, \mathrm{d}x\\ &=\frac{\pi ^{4}}{30}+\frac{1}{2}\left [\mathrm{Li}_2\left ( x \right ) \right ]^{2}\Bigg|_{0}^{1}\\ &=\frac{\pi ^{4}}{30}+\frac{\pi ^{4}}{72}\\ &=\frac{17\pi ^{4}}{360} \end{align*}