If a Dirichlet series converges Conditionally, how can I apply Euler product?

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?

• Apply the method for $s>1$ and let $s\to 1$ should work, though this will need some justification. (I don't really like the way you wrote the final formula, it almost looks as if two infinite products were taken.) Dec 11 '16 at 6:10
• @ChristianRemling that is more subtle for products than sums. There is not a simple version of Abel's theorem for infinite products. See Examples 3.5 and 5.13 in math.uconn.edu/~kconrad/articles/eulerprod.pdf. Dec 11 '16 at 14:01
• @ChristianRemling I cleaned up the last formula so it is not written like divergent products anymore. Dec 11 '16 at 14:16
• Two related MO questions that you might enjoy and benefit from: mathoverflow.net/questions/63714/… and mathoverflow.net/questions/63787/… Dec 11 '16 at 14:43

You are right to question this. The product $\prod_p \left(1 - \chi(p)/p\right)^{-1}$ (where $\chi = (-1/\cdot)$ is the Dirichlet character mod $4$) does converge, and the limit is $L(1,\chi) = \pi/4$ as expected; But this requires justification $-$ indeed it is equivalent to the non-vanishing of the Dirichlet function $L(s,\chi)$ on the edge $s = 1+it$ of the critical strip, which is also what you need to prove the analogue of the Prime Number Theorem for primes in arithmetic progressions mod $4$. (Taking logarithms, we see that $\prod_p \left(1 - \chi(p)/p\right)^{-1}$ converges if and only if $\sum_p \chi(p)/p$ converges, since this sum differs from the product's logarithm by an absolutely convergent sum $\sum_p O(1/p^2)$; getting from $\sum_p \chi(p)/p$ to $L(s,\chi)$, and then showing that the product $\prod_p \left(1 - \chi(p)/p\right)^{-1}$ actually converges to $L(1,\chi)$, is a classical chapter of analytic number theory.)
• Yes, but this is not true for every Euler product. The Dirichlet L-functions $L(s,\chi)$ are special because their coefficients are periodic, so they are entire and have a functional equation, from which we know the vertical density of zeros $\beta$ in the critical strip, allowing us to write $\frac{L'(s,\chi)}{L(s,\chi)} = -\sum_\beta \frac{1}{s-\beta}$, from which we get the explicit formula $\sum_{p^k < x} \frac{\chi(p^k)}{k} = \sum_\beta li(x^{\beta})+\mathcal{O}(1)$, Dec 11 '16 at 18:45
• so that if $L(s,\chi)$ has no zeros for $Re(s) >\sigma$ then $\sum_{p^k < x} \frac{\chi(p^k)}{k} = \mathcal{O}(x^{\sigma+\epsilon})$ and $\log L(s,\chi) = s \int_1^\infty (\sum_{p^k < x} \frac{\chi(p^k)}{k}) x^{-s-1}dx$ converges for $Re(s) >\sigma$. Dec 11 '16 at 18:46