TITLE: $1+\frac {1}{4}(1+\frac {1}{4}) +\frac {1}{9}(1+\frac {1}{4} +\frac {1}{9})+....$
QUESTION [8 upvotes]: Show that $$1+\frac {1}{4} \bigg(1+\frac {1}{4}\bigg) +\frac {1}{9} \bigg(1+\frac {1}{4} +\frac {1}{9}\bigg)+.....$$
converges.
Can you find the exact value of the sum.
My effort:
I have proved the convergence with comparing to $$\bigg(\sum _1^\infty \frac {1}{n^2}\bigg)^2$$
I have not figure out the exact sum.
Any suggestions??

REPLY [3 votes]: 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 $\displaystyle a_i={1\over i^2}$.