In the context of the Dirichlet Prime Theorem, we come to the equation
$$\sum_{p\equiv a \text{ mod } q} p^{-s} = \varphi(q)^{-1}\sum_{p \not\mid q}p^{-s} + \varphi(q)^{-1}\sum_{\chi\not=\chi_0}\bar{\chi}(a) \sum_p\chi(p)p^{-s}$$
Now it is argued that the first term on the right side diverges to $\infty$ for $s\to1$ and the second term is bounded. Therefore, the left side of above equation trends to $\infty$.
However, it then follows that $$\sum_{p\equiv a \text{ mod } q} p^{-1}$$ diverges and therefore, there are infinite primes in each arithmetic progression for coprime $a,q$.
But in order to make this conclusion, I have to argue that $$\lim_{s\to1}\sum_{p\equiv a \text{ mod } q} p^{-s} = \sum_{p\equiv a \text{ mod } q} p^{-1}$$
right? All the interchanging theorems that I know require the convergence of a series in the first place. But how do I prove the above equation knowing that the sum diverges? Or is there any other trivial argument why the divergence of $$\lim_{s\to1}\sum_{p\equiv a \text{ mod } q} p^{-s}$$ implies the one of $$\sum_{p\equiv a \text{ mod } q} p^{-1}$$
I would appreciate any help on this!
This works because the terms of the sequence are nonnegative and monotonically increase as $s\to 1$. Since $\sum_{p\equiv a \text{ mod } q} p^{-s}$ goes to $\infty$, for any $N$, there exists $s>1$ and $n$ such that $$\sum_{p\equiv a \text{ mod } q,p\leq n} p^{-s}\geq N.$$ Then $$\sum_{p\equiv a \text{ mod } q,p\leq n} p^{-1}\geq N$$ as well. Since the terms $p^{-1}$ for $p>n$ are all positive, this implies the infinite sum $\sum_{p\equiv a \text{ mod } q} p^{-1}$ is also at least $N$. Since $N$ was arbitrary, this means the infinite sum diverges.
(More generally, limits and sums can be interchanged when the limits are monotone increasing and the terms are nonnegative, even if infinite values are allowed. Even more generally, the same applies to limits and integrals, by the Lebesgue monotone convergence theorem.)