# What is the closed form of this sum?

What is the closed form of this sum? $$S = \sum_{k\ge1, r>s\ge 1}\frac{1}{k^2(r^2 + s^2)^2}$$

Note: Though originally posted for Pythagorean triplets, their was an flaw in the question which changed the meaning of the question and was answered accordingly. I will post a separate question with the original question on Pythagorean triangles.

Update: I have posted the related question of Pythagorean triangles in the link belwo

What is the sum of the reciprocal of the square of hypotenuse of Pythagorean triangles?

• @PaulSinclair Anything with a square root is not a Pythagorean triangle where all sides must be natural numbers. – Nilotpal Kanti Sinha Oct 21 at 17:00
• This $S$ is an interesting sum. But it is unclear what it should have to do with the geometric explanation given in the question. – Christian Blatter Oct 21 at 18:34
• @ChristianBlatter Actually I came up with this sum geometrically while wondering what values of the sum of the reciprocal of hypotenuse squares would converge to so to be fair, without the geometrical interpretation, I would have no reason to think about this sum in the first place. – Nilotpal Kanti Sinha Oct 21 at 19:05
• @NilotpalKantiSinha - Sorry, I was thinking just "right triangle". But that just means I picked a poor example. The problem I'm pointing out is still there. So a better example would be the triangle $(12, 16, 20)$, which is included for both $k = 1, r = 16, s = 12$ and for $k = 4, r =4, s = 3$. You need to restrict to $r, s$ being relatively prime, and one or the other should be even. – Paul Sinclair Oct 21 at 23:08
• @Blue Fair point. I will post a separate question for Pythagorean triplets and make this question for the sum $S$ as defined in the original question. – Nilotpal Kanti Sinha Oct 22 at 17:50

\begin{align*} S &= \sum_{k \geq 1} \frac{1}{k^2} \sum_{r>s\geq 1} \frac{1}{(r^2 + s^2)^2} \\ &= \frac{\pi^2}{6} \sum_{r>s\geq 1} \frac{1}{(r^2 + s^2)^2} \end{align*}

Note that $$\frac{1}{(r^2 + s^2)^2}$$ is unchanged by $$\{r \mapsto -r\}$$, $$\{s \mapsto -s\}$$, and $$\{r \leftrightarrow s\}$$. Applying these symmetries, we obtain eight copies of the region $$r > s \geq 1$$ in the $$r$$-$$s$$ plane. It lacks the four diagonals, each with value $$\displaystyle \sum_{s \geq 1} \frac{1}{(s^2 + s^2)^2} = \frac{\pi^4}{360}$$, the four coordinate rays starting at $$\pm 1$$, each with value $$\displaystyle \sum_{s \geq 1} \frac{1}{(0^2 + s^2)^2} = \frac{\pi^4}{90}$$, and the origin, which we will continue to exclude. Placing a prime on the summation sign to indicate that we omit the origin, $$\sideset{}{'}\sum_{r,s = -\infty}^\infty \frac{1}{(r^2 + s^2)^2} = 8 \sum_{r > s \geq 1} \frac{1}{(r^2 + s^2)^2} + 4 \cdot \frac{\pi^4}{360} + 4 \cdot \frac{\pi^4}{90}$$ Borwein and Borwein [1] (p. 291) (see also (38) at Double Series on MathWorld) show $$\sideset{}{'}\sum_{r,s = -\infty}^\infty \frac{1}{(r^2 + s^2)^2} = 4 \beta(2) \zeta(2) = \frac{2}{3} \pi^2 K \text{,}$$ where $$\beta$$ is Dirichlet's Beta function, $$\zeta$$ is the Riemann Zeta function, and $$K$$ is Catalan's constant.

Utilizing these facts, $$S = \frac{\pi^2}{6} \cdot \frac{12 K \pi^2 - \pi^4}{144} = 0.1264945807\dots \text{.}$$

[1] Borwein, Jonathan M. and Peter B. Borwein, "Pi and the AGM: A Study in Analytic Number Theory and Computational Complexity", New York: Wiley, 1987.

P.S. Mathematica 11.3 does not evaluate the triple sum symbolically. It numerically estimates $$0.126494$$, which is comfortably close to the above result.

• @NilotpalKantiSinha : Of course. The sum would be smaller, approximately $0.11687$. – Eric Towers Oct 22 at 14:33
• @NilotpalKantiSinha - you also have to require that either $r$ or $s$ is even. If they are both odd, then all three sides of the triangle will be divisible by $2$, which will duplicate another triangle in the sum. IIRC, every pythagorean triple is uniquely representable by $(k(r^2-s^2), 2krs, k(r^2 + s^2))$ with $\gcd(r,s) = 1$ and one of $r,s$ even. – Paul Sinclair Oct 22 at 16:21
• @NilotpalKantiSinha : Well, I don't currently see how to do the sum with the constraint $\gcd(r,s) = 1$, nor for Paul Sinclair's more complete set of constraints "$\gcd(r,s) = 1$ and one of $r,s$ is even". – Eric Towers Oct 22 at 16:22
• @NilotpalKantiSinha : With $\gcd(r,s) = 1$ and one of $r,s$ is even, the sum seems to be near $0.0935$ (and I don't trust the last digit). – Eric Towers Oct 22 at 16:36
• @NilotpalKantiSinha : I may have an idea for the restricted sum, but it will have to wait until tomorrow. – Eric Towers Oct 22 at 21:45