Find the sum of the infinite series given

Question

I've this problem to solve, but no answer key, I tried to do some development, but i didn't succeed :( Could anyone help me, please?

$$\sum_{k=2}^\infty \sum_{l=1}^{k-1} \frac{1}{k^2l^2}=\frac{1}{2^21^2}+\frac{1}{3^21^2}+\frac{1}{3^22^2}+\frac{1}{4^21^2}+\frac{1}{4^22^2}+\frac{1}{4^23^2}+\cdots$$

Hint: Use that $$\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$$ and $$\sum_{n=1}^\infty \frac{1}{n^4}=\frac{\pi^4}{90}$$ Thanks a lot!

Answer 1

Hint:

$$\left(\sum_{i=1}^na_i\right)^2=\left(\sum_{i=1}^na_i^2\right)+2\left(\sum_{0\leq i<j\leq n}a_ia_j\right)$$

Answer 2

Denoting the given double sum by $S$, we note that $$\left(\sum_{n=1}^\infty\frac1{n^2}\right)^2=\sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{m^2n^2}=\sum_{n=1}^\infty\frac1{n^4}+2S$$ Thus $$\left(\frac{\pi^2}6\right)^2=\frac{\pi^4}{90}+2S$$ $$S=\frac{\pi^4}{120}$$