A conjectural infinite series for $\frac{\pi}{2}$

Question

Can you provide a proof for the claim given below?

In this Wikipedia article the constant $\pi$ is represented by the following infinite series:

$$\pi=\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{\epsilon(n)}}{n}$$ where $\epsilon(n)$ is the number of prime factors of the form $p \equiv 1 \pmod{4}$ of $n$ . (Euler, 1748)

Similarly, we can formulate the following claim:

$$\frac{\pi}{2}=\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{\kappa(n)}}{n}$$ where $\kappa(n)$ is the number of prime factors of the form $p \equiv 3 \pmod{4}$ of $n$ .

The SageMath cell that demonstrates this claim can be found here.

Now, I don't know how to start the proof. Any hints or references are welcomed.

Answer 1

Allow me a small change in notation: $\Omega_1(n)$ for the number of prime factors of the form $4k+1$ (counted with multiplicity), $\Omega_3(n)$ for the number of prime factors of the form $4k+3$ (counted with multiplicity). $(-1)^{\Omega_k}$ is a multiplicative function, hence for any $s>1$ Euler's product gives

$$ L((-1)^{\Omega_k},s)=\sum_{n\geq 1}\frac{(-1)^{\Omega_k(n)}}{n^s}=\prod_{p\in\mathcal{P}}\left(1+\frac{(-1)^{\Omega_k(p)}}{p^s}+\frac{(-1)^{\Omega_k(p^{2})}}{p^{2s}}+\frac{(-1)^{\Omega_k(p^{3})}}{p^{3s}}+\ldots\right). $$

By separating the primes according to their residue class $(\text{mod }4)$ we have that the RHS equals

$$ \frac{1}{1-2^{-s}}\prod_{p\equiv k\pmod{4}}\frac{1}{1+p^{-s}}\prod_{p\equiv -k\pmod{4}}\frac{1}{1-p^{-s}}. $$

Now, let us consider the trivial and the non-trivial character $(\text{mod }4)$ together with their Dirichlet series:

$$ L(\chi_0,s)=\frac{\zeta(s)+\eta(s)}{2}=\sum_{k\geq 0}\frac{1}{(2k+1)^s} = \zeta(s)\left(1-\frac{1}{2^s}\right)=\prod_{p\neq 2}\frac{1}{1-p^{-s}}$$

$$ L(\chi_1,s)=\beta(s)=\sum_{k\geq 0}\frac{(-1)^k}{(2k+1)^s} = \prod_{p\equiv 1\pmod{4}}\frac{1}{1-p^{-s}}\prod_{p\equiv 3\pmod{4}}\frac{1}{1+p^{-s}}$$

$\beta(1)=\frac{\pi}{4}$ immediately proves both claims, since

$$\prod_{p\neq 2}\frac{1}{1+p^{-s}}=\prod_{p\neq 2}\frac{1-p^{-s}}{1-p^{-2s}}=\frac{1-2^{-2s}}{1-2^{-s}}\cdot\frac{\zeta(2s)}{\zeta(s)} $$

Answer 2

I believe A034947: Jacobi (or Kronecker) symbol (-1/n). provides some insight where $\beta(s)$ is the Dirichlet beta function:

$$(-1)^{\kappa(n)}=A034947(n)\tag{1}$$

$$\frac{\beta(s)}{1-2^{-s}}=\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N\frac{(-1)^{\kappa(n)}}{n^s}\right),\quad\Re(s)>0\tag{2}$$

$$\underset{s\to 1}{\text{lim}}\left(\frac{\beta(s)}{1-2^{-s}}\right)=\frac{\pi }{2}\tag{3}$$