Question about the distribution of primes in Davenport's Multiplicative Number Theory

Question

The following comes from Page 57 of Davenport's Multiplicative Number Theory. My question is: why is (6) being convergent necessary in deducing the result from (5)?

Answer 1

Have you studied the proof of Dirichlet's theorem enough to appreciate how a sum over the primes subject to a congruence condition $a \bmod q$, where $\gcd(a,q) = 1$, can be broken up into a linear combination over all Dirichlet characters mod $q$ of sums over all primes, where the term associated to a prime $p$ is weighted by a factor $\chi(p)$? This is what Davenport means by "a linear combination of characters in the usual way".

Here is how this idea arises in the proof of Dirichlet's theorem. When $s > 1$, $$ \sum_{p\equiv a \bmod q} \frac{1}{p^s} = \sum_p \frac{1}{\varphi(q)}\left(\sum_{\chi \bmod q} \overline{\chi}(a)\chi(p)\right)\frac{1}{p^s} = \frac{1}{\varphi(q)}\sum_{\chi \bmod q}\overline{\chi}(a)\sum_p \frac{\chi(p)}{p^s} $$ and you then separately look at the inner sum on the right where $\chi$ is trivial and the inner sums where $\chi$ is nontrivial: the inner sums at nontrivial $\chi$ are bounded as $s \to 1^+$, so $$ \sum_{p\equiv a \bmod q} \frac{1}{p^s} = \frac{1}{\varphi(q)}\sum_p \frac{1}{p^s} + O(1) $$ as $s \to 1^+$. This expresses a sum over $p$ in a congruence class mod $q$ as a sum over all $p$ plus an error term that is $O(1)$ because we know the sums over nontrivial $\chi \bmod q$ are bounded as $s \to 1^+$. Understanding the behavior of $\sum_p 1/p^s$ as $s \to 1^+$ then leads to Dirichlet's theorem in terms of Dirichlet density.

With more work, it turns out that each of the infinite series $\sum \chi(p)/p^s$ for nontrivial $\chi \bmod q$ is not just bounded as $s \to 1^+$, but it actually converges as $s \to 1^+$, say to $C_\chi$, and using this stronger information lets us refine the $O(1)$ error term in the above equation so we can say
$$ \sum_{p\equiv a \bmod q} \frac{1}{p^s} = \frac{1}{\varphi(q)}\sum_p \frac{1}{p^s} + C_{a,q} + o(1) $$ as $s \to 1^+$, where $C_{a,q} = (1/\varphi(q))\sum_{\chi \not= {\rm 1}_q} \overline{\chi}(a)C_\chi$, a constant depending on $a$ and $q$.

You should adapt this to the question you asked: there is no variable $s$ anymore, and now the infinite series over all primes subject to a congruence condition is replaced by a finite sum over the primes $p \leq x$ subject to a congruence condition. Express that finite sum as a "linear combination in the usual way" of the finite sums $\sum_{p \leq x} \chi(p)/p$ where $\chi$ runs over all Dirichlet characters mod $q$. Then separately consider such sums at trivial $\chi$ using (5) and at all nontrivial $\chi$ using (6).