TITLE: Triple Euler sum result $\sum_{k\geq 1}\frac{H_k^{(2)}H_k }{k^2}=\zeta(2)\zeta(3)+\zeta(5)$
QUESTION [44 upvotes]: In the following thread
I arrived at the following result 
$$\sum_{k\geq 1}\frac{H_k^{(2)}H_k }{k^2}=\zeta(2)\zeta(3)+\zeta(5)$$
Defining 
$$H_k^{(p)}=\sum_{n=1}^k \frac{1}{n^p},\,\,\, H_k^{(1)}\equiv H_k $$
But, it was after long evaluations and considering many variations of product of polylogarithm integrals.
I think there is an easier approach to get the solution, any ideas ?

REPLY [2 votes]: Lets start with the following double sum
$$\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{H_{n-1}}{n^2(n+k)^2}=\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{H_{n-1}}{n^2}\left(-\int_0^1 x^{n+k-1}\ln xdx\right)$$
$$=-\int_0^1 \ln x\left(\sum_{k=1}^\infty x^{k-1}\right)\left(\sum_{n=1}^\infty\frac{H_{n-1}}{n^2}x^n\right)dx$$
$$=-\int_0^1 \ln x\left(\frac{1}{1-x}\right)\left(\frac12\int_0^x \frac{\ln^2(1-y)}{y}dy\right)dx$$
$$=-\frac12\int_0^1 \frac{\ln^2(1-y)}{y}\left(\int_y^1\frac{\ln x}{1-x}dx\right)dy$$
$$=\frac12\int_0^1 \frac{\ln^2(1-y)\text{Li}_2(1-y)}{y}dy\overset{1-y=x}{=}\frac12\int_0^1 \frac{\ln^2x\text{Li}_2(x)}{1-x}dx$$
$$=\frac12\sum_{n=1}^\infty H_n^{(2)}\int_0^1 x^n \ln^2xdx=\sum_{n=1}^\infty\frac{H_n^{(2)}}{(n+1)^3}=\sum_{n=1}^\infty\frac{H_{n-1}^{(2)}}{n^3}=\sum_{n=1}^\infty\frac{H_{n}^{(2)}}{n^3}-\zeta(5)\tag1$$
On the other hand and by writing $\sum_{k=1}^\infty\frac{1}{(n+k)^2}=\zeta(2)-H_n^{(2)}$ we have
$$\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{H_{n-1}}{n^2(n+k)^2}=\sum_{n=1}^\infty\frac{H_{n-1}}{n^2}\left(\zeta(2)-H_n^{(2)}\right)$$
$$=\zeta(2)\sum_{n=1}^\infty\frac{H_{n-1}}{n^2}-\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^2}+\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}\tag2$$
$(1)-(2)$ gives us
$$\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^2}=\zeta(5)+\zeta(2)\sum_{n=1}^\infty\frac{H_{n-1}}{n^2}=\zeta(5)+\zeta(2)\zeta(3)$$
where the last sum follows from dividing both sides of $\sum_{n=1}^\infty\frac{H_{n-1}}{n}x^n=\frac12\ln^2(1-x)$ by $x$ then $\int_0^1$ , i.e
$$\sum_{n=1}^\infty\frac{H_{n-1}}{n^2}=\frac12\int_0^1\frac{\ln^2(1-x)}{x}dx=\frac12\int_0^1\frac{\ln^2x}{1-x}dx=\zeta(3)$$