In 1737, Euler discovered that if $ f(n) $ is multiplicative and $ \sum f(n)/n^{s} $ converges absolutely for ${\rm Re}(s) > \sigma_a$ then 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^2)}{p^{2s}}+ \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}}~~~~~{\rm if} ~~{\rm 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},~~~~~~~~{\rm where}~~~~f(n)=\begin{cases}(-1)^{(n-1)/2} & {\rm if} \ n \ {\rm odd}, \\ 0 & {\rm if } \ n \ {\rm even}, \end{cases} \end{equation} so the theorem gives \begin{equation} \frac{\pi}{4}~=~ \prod_{p \not= 2} \frac{1}{1-f(p)/p} ~=~ \prod_{p\not=2} \frac{p}{p-(-1)^{(p-1)/2}}~=~\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?