In this post, we will write $M_q$ as a set of Dirichlet characters modulo $q\geq 1$.
Theorem If $\chi\in M_q$ and $\chi\neq \chi_0$, then $L(1,\chi)\neq 0$.
Corollary: If $a$, $q$ are positive integers with $(a,q) = 1$, then there are infinitely many prime numbers $p$ such that $p\equiv a\pmod q$.
The problem is about the corollary. The lecturer proved it by proving that $\sum_{p\equiv a\pmod q}\frac{1}{p^s}$ diverges as $s\to 1$. The first question is, why are they equivalent?
The lecturer has shown that, for $\Re s>1$, $$\sum_{p\equiv a\pmod q}\frac{1}{p^s}=\frac{1}{\phi(q)}\sum_{\chi\in M_q}\overline{\chi}(a)\left ( \log L(s,\chi)+\mathcal{O}(1) \right )\tag{1}$$ He didn't go further, and let us do the rest ourselves. I hope you can correct me if I am mistaken somewhere. If $\chi=\chi_0$, then $$ (s-1)L(s,\chi_0)=(s-1)\zeta(s)\prod_{p\mid q}(1-p^{-s})\xrightarrow{s\to 1^{+}} \prod_{p\mid q}(1-p^{-1}) $$ The limit is finite and is $\neq 0$. This implies that $L(s,\chi_0)\xrightarrow{s\to 1^{+}} \infty$, and so $\log L(s,\chi_0)\xrightarrow{s\to 1^{+}} \infty$. The right hand side of $(1)$ is then divergent as $s\to 1^+$, which makes the left hand side divergent. If $\chi\neq \chi_0$, then $\log L(s,\chi)\xrightarrow{s\to 1^{+}} \log L(1,\chi)$ is finite by theorem above. How do I then check if the right hand side of $(1)$ diverges? This is the second question.