How to calculate $\sum_{n \in P}\frac{1}{n^2}, P=\{n \in \mathbb{N}: \exists (a,b) \in\ \mathbb{N^+} \times \mathbb{N^+} \mbox{ with } a^2+b^2=n^2\}$

Question

How I can evaluate $$\sum_{n \in P}\frac{1}{n^2} \quad \quad P=\{n \in \mathbb{N^+}: \exists (a,b) \in\ \mathbb{N^+} \times \mathbb{N^+} \mbox{ with } a^2+b^2=n^2\}$$ It's clearly convergent. I thought about seeing the sum as a sum of complex numbers using $(a+ib)(a-ib)=a^2+b^2$.

Answer 1

We just have to understand which numbers $n\in\mathbb{N}^+$ are such that $n^2$ cannot be written as $a^2+b^2$ with $a,b\in\mathbb{N}^+$. If there is some prime $p\equiv 1\pmod{4}$ that divides $n$, such prime splits in $\mathbb{Z}[i]$ (the ring of gaussian integers), hence $n^2$ can be represented for sure as $a^2+b^2$ with $a,b\neq 0$. If $n$ is even then $n^2\equiv 0\pmod{4}$, hence $n^2=a^2+b^2$ gives that both $a$ and $b$ are even and the problem boils down to representing $(n/2)^2$. At last we have that the only numbers whose square cannot be represented as $a^2+b^2$ with $a,b\neq 0$ are the ones whose prime factors lie in the set made by $2$ and the primes $\equiv 3\pmod{4}$. That gives: $$ \sum_{n\in P}\frac{1}{n^2} = \frac{\pi^2}{6}-\!\!\!\!\prod_{p\in\{2,3,7,\ldots\}}\left(1-\frac{1}{p^2}\right)^{-1}=\frac{\pi^2}{6}-\frac{4}{3}\cdot\!\!\!\prod_{p\equiv 3\!\!\pmod{4}}\left(1-\frac{1}{p^2}\right)^{-1}.\tag{1}$$ Now, let $\chi$ be the non-principal Dirichlet character $\!\!\pmod{4}$ and $L(s,\chi)$ the associated Dirichlet L-function. We have: $$ L(2,\chi) = \sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^2} = \prod_{p}\left(1-\frac{\chi(p)}{p^2}\right)^{-1},\tag{2} $$ $$ L(4,\chi) = \sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^4} = \prod_{p}\left(1-\frac{\chi(p)}{p^4}\right)^{-1},\tag{3} $$ $$ \frac{L(4,\chi)}{L(2,\chi)}=\prod_{p}\left(1+\frac{\chi(p)}{p^2}\right)^{-1}=\prod_{p\equiv 3\!\!\pmod{4}}\left(1-\frac{1}{p^2}\right)^{-1}\prod_{p\equiv 1\!\!\pmod{4}}\left(1+\frac{1}{p^2}\right)^{-1}\tag{4}$$ hence:

$$ S= \sum_{n\in P}\frac{1}{n^2}=\frac{\pi^2}{6}-\frac{4\, L(4,\chi)}{3\, L(2,\chi)}\cdot\!\!\prod_{p\equiv 1\!\!\pmod{\!\! 4}}\!\!\left(1+\frac{1}{p^2}\right).\tag{5}$$

Answer 2

Let $S(x)$ denote the number of positive integers not exceeding $x$ which can be expressed as a sum of two squares. Then, as proved by Landau in 1908 the following limit holds. $$ \lim_{x\to\infty} \frac{\sqrt{\ln x}}{x} S(x) = K, $$ where $K \approx 0.76422365358922$ is a constant. The convergence to the constant $K$, known as the Landau–Ramanujan constant is very slow. The first ten thousand digits of $K$ is here.

The exact value of $K$ can be expressed as $$ K = \frac{1}{\sqrt{2}} \prod_{\substack{p \text{ prime } \\ \equiv \, 3 \pmod{4}}} \left(1-\frac{1}{p^2}\right)^{-1/2}, $$ so we have that $$ \prod_{\substack{p \text{ prime } \\ \equiv \, 3 \pmod{4}}} \left(1-\frac{1}{p^2}\right) = \frac{1}{2K^2}. $$

From the excellent answer of @Jack D'Aurizio we know that $$ S = \sum_{n\in P}\frac{1}{n^2} = \frac{\pi^2}{6}-\frac{4}{3}\cdot \prod_{\substack{p \text{ prime } \\ \equiv \, 3 \pmod{4}}} \left(1-\frac{1}{p^2}\right)^{-1}. \\ $$

Using the exact value of $K$ we could express your sum in term of the Landau–Ramanujan constant.

$$ S = \sum_{n\in P}\frac{1}{n^2} = \frac{\pi^2}{6}-\frac{8K^2}{3}. $$

Numerically

$$ \begin{align} S \approx 0.08749995296754071824615285056063798739937787259940111394813\\ 9987884926591232919611579752270225245544983905801979851301833\\ 539947996701320476568130203602037004645936371\dots\phantom{0000000000000} \end{align} $$

Note that the numerical evaluation of the product for $K$ is hopeless in this form. You could find an other expression for $K$ here (eq. 11), which has much faster convergence.