Is $\sum_{p\text{ prime, } p \geq 2}\frac{(-1)^{\frac{p^2-1}{8}}}{p}$ convergent or divergent??

Question

Is $\sum\limits_{\text{p prime}, p \geq 2}\frac{(-1)^{\frac{p^2-1}{8}}}{p}$ convergent or divergent?

So far I have that

\begin{align} \sum\limits_{\text{p prime}, p \geq 2} \frac{(-1)^{\frac{p^2-1}{8}}}{p} =& \frac{(-1)^{\frac{3}{8}}}{2} + \sum\limits_{\text{p prime}, p > 2} \frac{(-1)^{\frac{p^2-1}{8}}}{p}\\ =& \frac{(-1)^{\frac{3}{8}}}{2} + \sum\limits_{\text{p prime}, p > 2}\left(\frac{2}{p}\right)\frac{1}{p}\\ =& \frac{(-1)^{\frac{3}{8}}}{2} + \sum\limits_{\substack{\text{p prime}, p > 2, \\ \text{ 2 a quadratic residue (mod p)}}}\frac{1}{p} \\ &- \sum\limits_{\substack{\text{p prime}, p > 2, \\ \text{ 2 a quadratic nonresidue (mod p)}}}\frac{1}{p}\\ \end{align}

I'm guessing these two sums should cancel - I remember a theorem of some sort which said that the number of quadratic residues 2(mod p) for all prime's $p > 2$ is $\sim \frac{\pi(x)}{2}$

Am I on the right track? or should I have done this differently? Any help and comments are appreciated.

Answer 1

Such a series behaves like $\log L(1,\chi)$ where $\chi$ is the Dirichlet character $\pmod{8}$ associated with the Legedre symbol $$\left(\frac{2}{p}\right)=(-1)^{\frac{p^2-1}{8}}, $$ hence the series is convergent.

We also have: $$\sum_{n\geq 1}\frac{\chi(n)}{n}=\sum_{n\geq 0}\left(\frac{1}{8n+1}-\frac{1}{8n+3}-\frac{1}{8n+5}+\frac{1}{8n+7}\right)=\frac{1}{2\sqrt 2}\log\frac{2+\sqrt{2}}{2-\sqrt{2}}.$$