Using integration, I managed to show that
$$\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{H_{n-1}}{n^2(n+k)^2}=\sum_{n=1}^\infty\frac{H_{n-1}^{(2)}}{n^3}\tag1$$
But I would like to prove the equality using series manipulation.
Here is what I did: $n+k=m$
$$\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{H_{n-1}}{n^2(n+k)^2}=\sum_{m=n+1}^\infty\sum_{n=1}^\infty\frac{H_{n-1}}{n^2m^2}$$
Next we use
$$\sum_{n=1}^\infty\sum_{m=n+1}^\infty=\sum_{m=1}^\infty\sum_{n=1}^{m-1}$$
$$\Longrightarrow \sum_{k=1}^\infty\sum_{n=1}^\infty\frac{H_{n-1}}{n^2(n+k)^2}=\sum_{m=1}^\infty\sum_{n=1}^{m-1}\frac{H_{n-1}}{n^2m^2}=\sum_{m=1}^\infty\frac{1}{m^2}\left(\sum_{n=1}^{m-1}\frac{H_{n-1}}{n^2}\right)\tag2$$
By Abel's summation I found
$$\sum_{n=1}^{m-1}\frac{H_{n-1}}{n^2}=H_m^{(2)}H_m-\frac{H_m}{m^2}+\frac1{m^3}-\sum_{n=1}^m\frac{H_n^{(2)}}{n}$$
Plugging this in $(2)$ does not give $(1)$. Any idea?
Thanks
Here is my integration work:
$$\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{H_{n-1}}{n^2(n+k)^2}=\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{H_{n-1}}{n^2}\left(-\int_0^1 x^{n+k-1}\ln xdx\right)$$
$$=-\int_0^1 \ln x\left(\sum_{k=1}^\infty x^{k-1}\right)\left(\sum_{n=1}^\infty\frac{H_{n-1}}{n^2}x^n\right)dx$$
$$=-\int_0^1 \ln x\left(\frac{1}{1-x}\right)\left(\frac12\int_0^x \frac{\ln^2(1-y)}{y}dy\right)dx$$
$$=-\frac12\int_0^1 \frac{\ln^2(1-y)}{y}\left(\int_y^1\frac{\ln x}{1-x}dx\right)dy$$
$$=\frac12\int_0^1 \frac{\ln^2(1-y)\text{Li}_2(1-y)}{y}dy\overset{1-y=x}{=}\frac12\int_0^1 \frac{\ln^2x\text{Li}_2(x)}{1-x}dx$$
$$=\frac12\sum_{n=1}^\infty H_n^{(2)}\int_0^1 x^n \ln^2xdx=\sum_{n=1}^\infty\frac{H_n^{(2)}}{(n+1)^3}=\sum_{n=1}^\infty\frac{H_{n-1}^{(2)}}{n^3}$$