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?
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.