I want to show that $$\lim_{s\to 1^+}\prod_{r \equiv 3 \mod 4} (1-r^{-s})^{-1} = \infty.$$
I showed that $$\lim_{s\to 1^+} \ (s-1) \prod _{q \equiv 1 \mod 4} (1-q^{-s})^{-1} = 1.$$ I feel that there is a simple step that I am overlooking to show that the result immediately follows. Any ideas on how to proceed/conclude? I'm hoping to deduce from this that the sum of reciprocals of primes congruent to 3 modulo 4 diverges.
Let $\chi_4$ be the non-principal Dirichlet's character $\!\!\pmod{4}$. For any $s>1$ we have
$$ L(\chi_4,s)=\sum_{n\geq 1}\frac{\chi_4(n)}{n^s} = \sum_{k\geq 0}\left(\frac{1}{(4k+1)^s}-\frac{1}{(4k+3)^s}\right)\\=\prod_{p\equiv 1\!\!\pmod{4}}\left(1-\frac{1}{p^s}\right)^{-1}\prod_{p\equiv 3\!\!\pmod{4}}\left(1+\frac{1}{p^s}\right)^{-1}=\zeta(s)\prod_{p\equiv 3\!\!\pmod{4}}\left(\frac{1+1/p^s}{1-1/p^s}\right)^{-1} $$ by Euler's product. As $s\to 1^+$, the LHS approaches $\frac{\pi}{4}$ and $\zeta(s)$ approaches $+\infty$ due to its simple pole at $s=1$ with residue $1$. It follows that $$ \lim_{s\to 1^+}\prod_{p\equiv 3\!\!\pmod{4}}\left(\frac{1-1/p^s}{1+1/p^s}\right)^{-1}=+\infty $$ implying $$ \lim_{s\to 1^+}\prod_{p\equiv 3\!\!\pmod{4}}\left(1-\frac{1}{p^s}\right)^{-1}=+\infty. $$