# $1+\frac {1}{4}(1+\frac {1}{4}) +\frac {1}{9}(1+\frac {1}{4} +\frac {1}{9})+…$

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??

• The sum is equal to $\sum_n\frac{\lceil d(n)/2\rceil}{n^2}$, if this helps. – ajotatxe Nov 15 '18 at 16:19
• According to Mathematica, $\sum _{n=1}^{\infty } \frac{\sum _{i=1}^n \frac{1}{i^2}}{n^2}={7\pi^4\over360}$. Maybe this gives you an idea of how to derive the sum. – Steve Kass Nov 15 '18 at 16:26
• @SteveKass It should be related to $\zeta(4)$ – ajotatxe Nov 15 '18 at 16:28
• @SteveKass Thanks for the comment, yes sometimes the answer helps to figure out a solution. – Mohammad Riazi-Kermani Nov 15 '18 at 16:28

$$2S = \sum_{i\leq j} \frac{1}{i^{2}j^{2}} + \sum_{i\geq j} \frac{1}{i^{2}j^{2}} = \left(\sum_{n\geq 1}\frac{1}{n^{2}}\right)^{2} + \sum_{n\geq 1}\frac{1}{n^{4}} = \frac{\pi^{4}}{36} + \frac{\pi^{4}}{90}$$
• We can generalize the result to $$\sum_{i\le j}\frac{1}{i^r j^r}=\frac12\left(\zeta(r)^2+\zeta(2r)\right).$$ – Tianlalu Nov 15 '18 at 16:43
Suppose that $$\sum_{i=0}^\infty a_i$$ is an absolutely convergent series. Then (where $$i$$ and $$j$$ range over the nonnegative integers)
$$2\sum_{i<=j} a_ia_j=\\ 2\sum_{ij} a_ia_j+\color{green}{2\sum_{i=j} a_ia_j}=\\ \left(\sum_{ij} a_ia_j\right)+\color{green}{\sum_{i=j} a_ia_j}=\\ \color{red}{{\sum_{i,j} a_ia_j}}+\color{blue}{\sum_{i=j} a_ia_j}=\\ \color{red}{\left(\sum_{i} a_i\right)^2}+\color{blue}{\sum_{i} (a_i)^2}.$$
The question here is answered by this identity for $$\displaystyle a_i={1\over i^2}$$.