Convergence of $\sum_{n\leq x} \frac{\chi(n) \Lambda(n)}{n}$ I am trying to prove convergence of certain series related to non-principal Dirichlet series. In the proof, I want to use the following fact:
$$ \sum_{n\leq x} \frac{\chi(n)\Lambda(n)}{n} \tag{1} $$
converges as $x\to\infty$. Here $\chi$ is some non-principal character (say, mod $k$) and $\Lambda$ is the von Mangoldt function.
The only proof of convergence of (1) I know follows from Lemmas 7.3-7.8 in Apostol's Introduction to Analytic Number Theory. The proof feels a little bit long-winded for me (maybe because I do not understand it well), so here is my question:

How would one directly prove that (1) converges ? 

 A: The convergence of $(1)$ lies relatively deep, and Apostol only proves the boundedness, not convergence, since that is what is needed for the version of Dirichlet's theorem proved in Apostol's book (the version and proof are due to Shapiro). The boundedness of $(1)$ is equivalent to the non-vanishing of $L(1,\chi)$. This is fairly easy to see if one concentrates on just that. Following Mertens (but using modern notation), we have
\begin{align}
-L'(1,\chi) + O\biggl(\frac{\log x}{x}\biggr)
&= \sum_{k \leqslant x} \frac{\chi(k)\log k}{k} \\
&= \sum_{k \leqslant x} \frac{\chi(k)}{k}\sum_{n \mid k} \Lambda(n) \\
&= \sum_{k \leqslant x} \sum_{n\cdot m = k} \frac{\chi(m)}{m}\cdot \frac{\chi(n)\Lambda(n)}{n} \\
&= \sum_{n \leqslant x} \frac{\chi(n)\Lambda(n)}{n} \sum_{m \leqslant x/n} \frac{\chi(m)}{m} \\
&= \sum_{n \leqslant x} \frac{\chi(n)\Lambda(n)}{n}\biggl(L(1,\chi) + O\biggl(\frac{n}{x}\biggr)\biggr) \\
&= L(1,\chi)\sum_{n \leqslant x} \frac{\chi(n)\Lambda(n)}{n} + O\biggl(\frac{1}{x} \sum_{n \leqslant x} \lvert \chi(n)\Lambda(n)\rvert\biggr) \\
&= L(1,\chi)\sum_{n \leqslant x} \frac{\chi(n)\Lambda(n)}{n} + O(1)\,.
\end{align}
The last step uses Chebyshev's bound $\psi(x) = O(x)$. Thus, if $L(1,\chi) \neq 0$ we can divide by that and rearrange to see that $(1)$ is bounded. For the converse direction we note that boundedness of $(1)$ implies boundedness of
$$\sum_{p \leqslant x} \frac{\chi(p)\log p}{p} \tag{2}$$
since
$$\Biggl\lvert\sum_{\substack{p^m \leqslant x \\ m \geqslant 2}} \frac{\chi(p^m)\log p}{p^m}\Biggr\rvert \leqslant \sum_{\substack{p^m \leqslant x \\ m \geqslant 2}} \frac{\log p}{p^m} < \sum_{\substack{p^m \\ m \geqslant 2}} \frac{\log p}{p^m} = \sum_p \frac{\log p}{p(p-1)} < +\infty\,,$$
and by Dirichlet's test boundedness of $(2)$ implies the convergence of
$$\sum_p \frac{\chi(p)}{p}\,.$$
This in turn implies the convergence of
$$\sum_p \log \biggl(1 - \frac{\chi(p)}{p}\biggr)$$
and hence
$$L(1,\chi) = \exp \bigl(-\sum_p \log \biggl(1 - \frac{\chi(p)}{p}\biggr)\biggr) \neq 0\,.$$
The convergence of $(1)$ is equivalent to the non-vanishing of $L(s,\chi)$ on the whole line $\sigma = \operatorname{Re} s = 1$, just like the convergence of
$$\sum_{n = 1}^{\infty} \frac{\mu(n)}{n}$$
is equivalent to the non-vanishing of $\zeta(s)$ on that line, and we can prove it in an analogous manner.
First we note for the implication "$(1)$ converges $\implies L(s,\chi) \neq 0$ for $\sigma = 1$" that generally the convergence of a Dirichlet series at a point $s_0 = \sigma_0 + i t_0$ implies that the function $F(s)$ represented by the Dirichlet series cannot have a pole on the line $\sigma = \sigma_0$, since for every $t \in \mathbb{R}$ we then have (Proposition 3)
$$\lim_{\varepsilon \searrow 0} \; \varepsilon \cdot F(\sigma_0 + \varepsilon + it) = 0\,.$$
But a zero of $L(s,\chi)$ is a pole of $-\frac{L'(s,\chi)}{L(s,\chi)}$.
For the other direction, that the absence of zeros on $\sigma = 1$ implies convergence of $(1)$, one can use many of the same techniques used to prove the prime number theorem. For example Newman's Tauberian theorem tells us (since we already know $(1)$ is bounded) that
$$\int_1^{\infty} \frac{1}{x} \biggl(\frac{L'(1,\chi)}{L(1,\chi)} + \sum_{n \leqslant x} \frac{\chi(n)\Lambda(n)}{n}\biggr)\,dx$$
exists as an improper integral. Since the sum changes slowly, it follows that the parenthesis in the integral must converge to $0$.
The proof that $L(s,\chi)$ has no zeros on $\sigma = 1$ can be modelled on Mertens's proof for the $\zeta$-function. If $\chi$ is a non-real character, or $t \neq 0$, then
$$L(\sigma,\chi_0)^3 \cdot \lvert L(\sigma + it, \chi)\rvert^4 \cdot \lvert L(\sigma + 2it, \chi^2)\rvert \geqslant 1 \tag{3}$$
for all $\sigma > 1$. Since $L(s,\chi_0)$ has a simple pole at $1$, this is only compatible with $L(1 + it, \chi) = 0$ if $L(s,\chi^2)$ has a pole at $1 + 2it$. But the only pole of all the $L$-functions is the pole of $L(s,\chi_0)$ at $1$, so that would mean $\chi^2 = \chi_0$, i.e. $\chi$ real, and $t = 0$. But this is the case excluded from consideration. (We see once again that the case of $L(1,\chi)$ for real characters is special; all other cases can be done at once with a single argument.)
One can use $(3)$ (or a similar inequality using a different cosine polynomial to obtain better estimates) and upper estimates for $\lvert L(s,\chi)\rvert$ and $\lvert L'(s,\chi)\rvert$ for $\sigma \geqslant 1 - \frac{c}{\log \lvert t\rvert}$ to obtain lower bounds for $\lvert L(s,\chi)\rvert$ in a region $\sigma \geqslant 1 - \frac{a}{(\log \lvert t\rvert)^{\alpha}}$ (the exponent $\alpha$ depends on the used cosine polynomial, for the classical Mertens polynomial $3 + 4\cos \varphi + \cos (2\varphi) = 2(1 + \cos \varphi)^2$ we have $\alpha = 9$, the "cubic" $5 + 8\cos \varphi + 4\cos (2\varphi) + \cos (3\varphi)$ gives $\alpha = 7$) and an $O\bigl(x\exp\bigl(-b(\log x)^{1/(\alpha+1)}\bigr)\bigr)$ error term in the prime number theorem for arithmetic progressions.
But the convergence of $(1)$ (for all non-principal characters modulo $k$) alone already gives
$$\sum_{\substack{n \leqslant x \\ n \equiv h \pmod{k}}} \frac{\Lambda(n)}{n}
= \frac{1}{\varphi(k)}\log x + C_{k,h} + o(1) \tag{4}$$
and consequently
$$\psi_{k,h}(x) = \sum_{\substack{n \leqslant x \\ n \equiv h \pmod{k}}} \Lambda(n)
= \frac{x}{\varphi(k)} + o(x)$$
and $\pi_{k,h}(x) \sim \frac{\operatorname{Li}(x)}{\varphi(k)}$ for $(h,k) = 1$, i.e. the PNT for the arithmetic progressions with difference $k$.
