Convergence of $\sum_{p>2} \frac{(-1)^{\frac{p-1}{2}}}{p}$

Question

Consider the sum $\sum_{p>2} \frac{(-1)^{\frac{p-1}{2}}}{p}$ where $p$ runs only through all odd primes. Show that this sum converges. The possibly best approach I have until now is via Partial summation, but dealing with number of primes is troubling, especially for obtaining explicit bounds.

Any help appreciated!

Answer 1

This is $\sum_p \frac{\chi(p)}{p}$, where $\chi$ denotes the quadratic Dirichlet character modulo $4$. We note that for $\Re(s) > 1$, $$\log L(s,\chi) = \log \prod_p \frac{1}{1 - \chi(p) p^{-s}} = -\sum_p \log (1 - \chi(p) p^{-s}) = \sum_p \sum_{k = 1}^{\infty} \frac{\chi(p)^k}{kp^{ks}}.$$ By the zero-free region for $L(s,\chi)$, this identity extends to $s = 1$. Furthermore, $$L(1,\chi) = \sum_{n = 1}^{\infty} \frac{\chi(n)}{n} = \frac{\pi}{4}$$ (either by Dirichlet's class number formula or by the power series expansion of $\arctan(x)$), and so $$\sum_p \frac{\chi(p)}{p} = \log \frac{\pi}{4} - \sum_{k = 2}^{\infty} \sum_{p} \frac{\chi(p)^k}{kp^k}.$$ There are various ways to bound the second term. For example, the contribution from the term for which $k = 2$ is $\frac{1}{8} - \frac{1}{2}\sum_p \frac{1}{p^2} \approx -0.101$ (by Wolfram Alpha, since $\sum_p p^{-s}$ is the prime zeta function). The remaining terms can be bounded by noting that $$\left|\sum_{k = 3}^{\infty} \sum_{p} \frac{\chi(p)^k}{kp^k}\right| < \frac{1}{3} \sum_{p > 2} \sum_{k = 3}^{\infty} \frac{1}{p^k} = \frac{1}{3} \sum_{p > 2} \frac{1}{p^2(p - 1)} < \frac{1}{3} \sum_p \frac{1}{p^3} \approx 0.058$$ (again using Wolfram Alpha for the last sum). With more effort, one can of course improve this.

Answer 2

Well, it depends on whether $p \bmod 4$ is 1 or 3.

What matters is $m(x, k, j) =\#\{\text{primes }kn+j\le x\} $ for $k=4, j = 1, 3 $.

It is known that, as $x \to \infty$, $\dfrac{m(x, 4, 3)}{m(x, 4, 1)} \to 1 $.

From the study of "primes races" (see https://dms.umontreal.ca/~andrew/PDF/PrimeRace.pdf ) it is known that $m(x, 4, 3)$ is usually larger that $m(x, 4, 1)$.

However, Littlewood showed that there are arbitrarily large $x$ such that $m(x, 4, 1)-m(x, 4, 3) \ge \dfrac{\sqrt{x}\ln\ln(x)}{2\ln(x)} $.

The following results are known.

$m(x, k, j) \sim \dfrac{x}{\phi(k)\ln(x)} \sim \dfrac{li(x)}{\phi(k)} $ (where $li(x)$ is the logarithmic integral and $\phi(k)$ is Euler's phi function) and, as referenced in https://primes.utm.edu/notes/Dirichlet.html,

$m(x, k, j) - \dfrac{li(x)}{\phi(k)} =O(xe^{-a\sqrt{\ln(x)}}) $ for $a = 1/15$.

This implies that $|m(x, 4, 3)-m(x, 4, 1)| =O(xe^{-a\sqrt{\ln(x)}}) $.

I believe that this is enough to show that the sum in the problem converges.

But I don't know for sure and I'll leave it at this.