# Evaluation of infinite sum

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!

• You may find the Wikipedia article multiple zeta function helpful. Commented May 14, 2020 at 17:26

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

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

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

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

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

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

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.

• This is the origin of my problem ;) I had to use this to calculate $\zeta (4)$. Commented May 15, 2020 at 8:58