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


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??

  • $\begingroup$ 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

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

  • 1
    $\begingroup$ @Seewood Lee Good answer, thanks. $\endgroup$ – Mohammad Riazi-Kermani Nov 15 '18 at 16:32
  • 1
    $\begingroup$ 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

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 $\displaystyle a_i={1\over i^2}$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.