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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??

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  • $\begingroup$ The sum is equal to $\sum_n\frac{\lceil d(n)/2\rceil}{n^2}$, if this helps. $\endgroup$ – ajotatxe Nov 15 '18 at 16:19
  • $\begingroup$ 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. $\endgroup$ – Steve Kass Nov 15 '18 at 16:26
  • $\begingroup$ @SteveKass It should be related to $\zeta(4)$ $\endgroup$ – ajotatxe Nov 15 '18 at 16:28
  • $\begingroup$ @SteveKass Thanks for the comment, yes sometimes the answer helps to figure out a solution. $\endgroup$ – Mohammad Riazi-Kermani Nov 15 '18 at 16:28
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$$ 2S = \sum_{i\leq j} \frac{1}{i^{2}j^{2}} + \sum_{i\geq j} \frac{1}{i^{2}j^{2}} = \left(\sum_{n\geq 1}\frac{1}{n^{2}}\right)^{2} + \sum_{n\geq 1}\frac{1}{n^{4}} = \frac{\pi^{4}}{36} + \frac{\pi^{4}}{90} $$

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  • 1
    $\begingroup$ @Seewood Lee Good answer, thanks. $\endgroup$ – Mohammad Riazi-Kermani Nov 15 '18 at 16:32
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    $\begingroup$ We can generalize the result to $$\sum_{i\le j}\frac{1}{i^r j^r}=\frac12\left(\zeta(r)^2+\zeta(2r)\right).$$ $\endgroup$ – Tianlalu Nov 15 '18 at 16:43
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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 $\displaystyle a_i={1\over i^2}$.

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  • $\begingroup$ Very beautiful, thank for the answer.. $\endgroup$ – Mohammad Riazi-Kermani Nov 15 '18 at 21:33

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