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?

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    $\begingroup$ The PNT in arithmetic progressions says that for $gcd(a,q)=1$ and any $r$ as $x \to \infty$, $ \pi(x;a \bmod q)=\sum_{p \le x,p\equiv a \bmod q} 1 = \frac{1}{\varphi(q)}\sum_{2 \le n \le x}\frac{1}{\log n} + O(\frac{x}{\log^r x})$. Then use partial summation to make $ \sum_{p \le x }\chi(p)$ appear from $\sum_{p \le x} \frac{\chi(p)}{p}$. $\endgroup$ – reuns Feb 15 at 1:56

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