Convergence of $\sum\limits_p \frac{\chi(p)}{p}$ and the prime number theorem Consider the sum
$$\sum\limits_p (-1)^{\frac{p-1}{2}}\frac{1}{p}=-\frac{1}{3} + \frac{1}{5} - \frac{1}{7} - \frac{1}{11} + \frac{1}{13} + \frac{1}{17} - \cdots \tag{1}$$
of signed reciprocals of the odd prime numbers, the sign being positive or negative according to whether $p$ is congruent to $1$ or $-1$ modulo $4$.  It is a consequence of Dirichlet's theorem on arithmetic progressions that the positive and negative terms are both divergent series, which implies that (1) is a conditionally convergent series.
It is supposedly a much deeper result that 
$$\lim\limits_{n \to \infty} -\frac{1}{3} + \frac{1}{5} - \cdots + \frac{(-1)^{\frac{p-1}{2}}}{p_n} \tag{2}$$
exists, where $p_n$ is the $n$th prime number.  User reuns explained something about this in the comments of my previous question, that is was actually equivalent to the prime number theorem.  I want to understand reuns' comment.   How can one show that the limit (2) exists, and what does this have to do with the prime number theorem?
 A: The (conditional) convergence of the series $\sum \frac{\chi(p)}{p}$, i.e. the existence of
$$\lim_{x \to \infty} \sum_{p \leqslant x} \frac{\chi(p)}{p}$$
for every non-principal Dirichlet character $\chi$ was proved by Mertens in his famous paper "Ein Beitrag zur analytischen Zahlentheorie" (Journal für die reine und angewandte Mathematik, 1874). The proof uses Dirichlet's result $L(1,\chi) \neq 0$, but not the prime number theorem.
If $\chi$ is a non-principal Dirichlet character, the Dirichlet series
$$L(s,\chi) = \sum_{n = 1}^{\infty} \frac{\chi(n)}{n^s} \qquad\text{and}\qquad L'(s,\chi) = \sum_{n = 1}^{\infty} \frac{-\chi(n)\log n}{n^s}$$
converge (conditionally) for $\operatorname{Re} s > 0$, in particular for $s = 1$. Since $\chi$ is completely multiplicative and
$$\log n = \sum_{d \mid n} \Lambda(d),$$
where $\Lambda$ is the von Mangoldt function,
$$\Lambda(n) = \begin{cases} \log p &\text{if } n = p^k \text{ with a prime } p \text{ and } k \geqslant 1,\\ \quad 0 &\text{otherwise,} \end{cases}$$
we have
$$\chi(n) \log n = \chi(n)\sum_{d \mid n} \Lambda(d) = \sum_{d\mid n} \chi(n)\Lambda(d) = \sum_{d\mid n} \chi(n/d)\cdot \chi(d)\Lambda(d).$$
Thus
\begin{align}
\sum_{n \leqslant x} \frac{\chi(n)\log n}{n}
&= \sum_{k\cdot m \leqslant x} \frac{\chi(k)\cdot \chi(m)\Lambda(m)}{k\cdot m} \\
&= \sum_{m \leqslant x} \frac{\chi(m)\Lambda(m)}{m} \sum_{k \leqslant x/m} \frac{\chi(k)}{k} \\
&= \sum_{m \leqslant x} \frac{\chi(m)\Lambda(m)}{m}\biggl(L(1,\chi) + O\Bigl(\frac{m}{x}\Bigr)\biggr) \\
&= L(1,\chi)\sum_{m \leqslant x} \frac{\chi(m)\Lambda(m)}{m} + O\biggl(\frac{1}{x}\sum_{m \leqslant x} \frac{m\lvert\chi(m)\rvert\Lambda(m)}{m}\biggr) \\
&= L(1,\chi)\sum_{m \leqslant x} \frac{\chi(m)\Lambda(m)}{m} + O(1),
\end{align}
where we have used $\sum_{m \leqslant x} \lvert\chi(m)\rvert\Lambda(m) \leqslant \psi(x) \in O(x)$ and $\bigl\lvert \sum_{k > y} \chi(k)/k\bigr\rvert \leqslant c/y$ for a constant $c$ depending only on $\chi$. Since $\sum \frac{\chi(n)\log n}{n}$ converges, its partial sums are uniformly bounded, so the above shows
$$L(1,\chi)\sum_{m \leqslant x} \frac{\chi(m)\Lambda(m)}{m} \in O(1).$$
Now using $L(1,\chi) \neq 0$ yields
$$\sum_{m \leqslant x} \frac{\chi(m)\Lambda(m)}{m} \in O(1).$$
Since
$$\sum_{\substack{m \\ m \text{ not prime}}} \frac{\Lambda(m)}{m} = \sum_{\substack{p^k \\ k \geqslant 2}} \frac{\log p}{p^k} = \sum_p \frac{\log p}{p(p-1)} < +\infty$$
it follows that also
$$\sum_{p \leqslant x} \frac{\chi(p)\log p}{p} \in O(1).$$
Then Dirichlet's criterion — $\frac{1}{\log p}$ converges to $0$ monotonically — yields the existence of
$$\lim_{x \to \infty} \sum_{p \leqslant x} \frac{\chi(p)}{p}.$$
This is essentially how Mertens proved the convergence, only the notation is more convenient and allows a more concise exposition nowadays.
