Evaluation of infinite sum

Question

In solving a problem, I stumbled upon this simple looking infinite sum: $$\sum\limits_{i=1}^\infty\sum\limits_{j=1}^i \frac{1}{i^2}\frac{1}{j^2}$$ At first it looked solvable, I wrote it down and noted that I could at least write it as: $$\zeta(4)+\zeta(2)+\sum\limits_{i=3}^\infty\sum\limits_{j=2}^{i-1}\frac{1}{i^2}\frac{1}{j^2}$$ Where $\zeta$ is the Riemann Zeta function.

With that sum, I got stuck. I tried to write down all the remaining "cross terms" in a neater way but this was not succesfull.

This really feels like something solvable. Any hints are really appreciated!

Answer 1

Swapping the order of the summation we get

$$S \equiv \sum_{j=1}^\infty \sum_{i=j}^\infty \frac{1}{i^2}\frac{1}{j^2} = \sum_{j=1}^\infty \sum_{i=1}^\infty \frac{1}{i^2}\frac{1}{j^2} - \sum_{j=1}^\infty \sum_{i=1}^{j-1} \frac{1}{i^2}\frac{1}{j^2}$$

Since the $i$'s and $j$'s are dummy variables, we can switch them in the last expression to get that

$$S = \sum_{j=1}^\infty \sum_{i=1}^\infty \frac{1}{i^2}\frac{1}{j^2} - \sum_{i=1}^\infty \sum_{j=1}^{i-1} \frac{1}{i^2}\frac{1}{j^2} = \sum_{j=1}^\infty \sum_{i=1}^\infty \frac{1}{i^2}\frac{1}{j^2} + \sum_{i=1}^\infty \frac{1}{i^4} - S$$

$$\implies 2S = \left(\sum_{i=1}^\infty \frac{1}{i^2}\right)^2 + \sum_{i=1}^\infty \frac{1}{i^4} = \frac{\pi^4}{36} + \frac{\pi^4}{90}$$

$$\therefore S = \frac{7\pi^4}{360}$$

Answer 2

Hint:

$$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} (\pi x)^{2n+1}=\sin \pi x=\pi x\prod_{n=1}^{\infty}\left( 1 - \frac{x^2}{n^2} \right)$$

Compare the coefficients of $x^3$ and $x^5$ on the far LHS to those on the far RHS.

Answer 3

Let $ n $ be a positive integer, we have : \begin{aligned}\left(\sum_{k=1}^{n}{\frac{1}{k^{2}}}\right)^{2}&=\sum_{1\leq i,j\leq n}{\frac{1}{i^{2}j^{2}}}\\ &=\sum_{1\leq i\leq j\leq n}{\frac{1}{i^{2}j^{2}}}+\sum_{1\leq j<i\leq n}{\frac{1}{i^{2}j^{2}}}\\ &=2\sum_{1\leq i\leq j\leq n}{\frac{1}{i^{2}j^{2}}}-\sum_{k=1}^{n}{\frac{1}{k^{4}}}\\ \left(\sum_{k=1}^{n}{\frac{1}{k^{2}}}\right)^{2}&=2\sum_{i=1}^{n}{\sum_{j=1}^{i}{\frac{1}{i^{2}j^{2}}}}-\sum_{k=1}^{n}{\frac{1}{k^{4}}}\end{aligned}

Thus, for any $ n\in\mathbb{N}^{*} $, we have : $$ \sum_{i=1}^{n}{\sum_{j=1}^{i}{\frac{1}{i^{2}j^{2}}}}=\frac{1}{2}\left(\sum_{k=1}^{n}{\frac{1}{k^{4}}}+\left(\sum_{k=1}^{n}{\frac{1}{k^{2}}}\right)^{2}\right) $$

Taking $ n $ to $ +\infty $, we get : $$ \sum_{i=1}^{+\infty}{\sum_{j=1}^{i}{\frac{1}{i^{2}j^{2}}}}=\frac{1}{2}\left(\zeta\left(4\right)+\zeta^{2}\left(2\right)\right)=\frac{7\pi^{4}}{360} $$