$1+\frac {1}{4}(1+\frac {1}{4}) +\frac {1}{9}(1+\frac {1}{4} +\frac {1}{9})+....$

$$ 2S = \sum_{i\leq j} \frac{1}{i^{2}j^{2}} + \sum_{i\geq j} \frac{1}{i^{2}j^{2}} = \left(\sum_{n\geq 1}\frac{1}{n^{2}}\right)^{2} + \sum_{n\geq 1}\frac{1}{n^{4}} = \frac{\pi^{4}}{36} + \frac{\pi^{4}}{90} $$

Suppose that $\sum_{i=0}^\infty a_i$ is an absolutely convergent series. Then (where $i$ and $j$ range over the nonnegative integers)

$$2\sum_{i<=j} a_ia_j=\\ 2\sum_{i<j} a_ia_j+2\sum_{i=j} a_ia_j=\\ \sum_{i<j} a_ia_j+\sum_{i>j} a_ia_j+\color{green}{2\sum_{i=j} a_ia_j}=\\ \left(\sum_{i<j} a_ia_j+\color{green}{\sum_{i=j} a_ia_j}+\sum_{i>j} a_ia_j\right)+\color{green}{\sum_{i=j} a_ia_j}=\\ \color{red}{{\sum_{i,j} a_ia_j}}+\color{blue}{\sum_{i=j} a_ia_j}=\\ \color{red}{\left(\sum_{i} a_i\right)^2}+\color{blue}{\sum_{i} (a_i)^2}.$$

The question here is answered by this identity for $\displaystyle a_i={1\over i^2}$.