Convergence of Euler Product for Leibniz Pi Formula

In class we were shown a derivation of Leibniz's formula for pi: $$\frac{\pi}{4}=\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}$$ We can rewrite this formula using the following function on $\mathbb{N}$: $$\chi(n)= \begin{cases}1, &\text{if n\equiv 1\pmod 4}\\ -1,& \text{if n\equiv 3\pmod 4}\\ 0, & \text{if n is even} \end{cases}$$ Leibniz's Formula then becomes $$\sum_{n=1}^\infty\frac{\chi(n)}{n}$$ We can rewrite this as an Euler product: $$\sum_{n=1}^\infty\frac{\chi(n)}{n}=\prod_{p\text{ prime}}\frac{1}{1-\chi(p)/p}$$ I get how the formula was obtained, but I'm not sure why this product converges to $\pi/4$. When I look up proofs for Euler products of Dirichlet series, I get results that assume that the series absolutely converges (which I don't think is true for Leibniz's pi formula). Any help would be greatly appreciated.

• $\prod_p \frac{1}{1-\chi(p)p^{-s}}$ converges to $\sum_{n=1}^\infty \chi(n) n^{-s}$ for $\Re(s) \ge 1$, this is equivalent to the prime number theorem for this L-function : $\sum_{p \le x} \chi(p) = o(\frac{x}{(\log x)^k})$. The (generalized) Riemann hypothesis is $\sum_{p \le x} \chi(p) = \mathcal{O}(x^{1/2}(\log x)^2)$ so that the product would converge for $\Re(s) > 1/2$. – reuns Jul 21 '17 at 2:24
• Are you asking about a proof of Leibniz's formula? – Klangen Dec 17 '18 at 15:05