In 1737, Euler discovered that if $ f(n) $ is multiplicative and $ \sum f(n)/n^{-s} $ converges absolutely for $ Re(s) > \sigma_a $ we have \begin{equation} \sum_{n=1}^{\infty} \frac{f(n)}{n^s} ~=~ \prod_p \Bigg\{ 1+\frac{f(p)}{p^s}+\frac{f(p)}{p^s}+ \cdots \Bigg\} \end{equation} and, especially, if $f$ is completely multiplicative we have \begin{equation} \sum_{n=1}^{\infty} \frac{f(n)}{n^s} ~=~ \prod_p \frac{1}{1-f(p)p^{-s}}~~~~~,if ~~Re(s)>\sigma_a~~~~. \end{equation}

I found an example in Wikipedia (https://en.wikipedia.org/wiki/Euler_product) like

\begin{equation} \frac{\pi}{4}~=~ \sum_{n=1}^{\infty} \frac{f(n)}{n}~~~~~~~,~where ~~~~f(n)=sin(n\pi/2) \end{equation} thus the theorem gives \begin{equation} \frac{\pi}{4}~=~ \Bigg( \prod_{4|(p-1)} \frac{p}{p-1} \Bigg) \Bigg( \prod_{4|(p-3)} \frac{p}{p+1} \Bigg)~=~\frac{3}{4}\cdot\frac{5}{4}\cdot\frac{7}{8}\cdot\frac{11}{12}\cdot \frac{13}{12}\cdots~~~~~~~~~~~~~. \end{equation} However, this example does Not converge absolutely but conditionally. If this example holds, how can I prove it though it converges conditionally? Is there any other additional condition needed or should I apply a different method?