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Show that the sum $$\sum \frac{(-1)^{\frac{p-1}{2}}}{p}$$ converges, where the sum is taken over all odd primes.

This problem was on an old Harvard qualifying exam. Is there a reasonably elementary way to solve it?

Potato
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We have that $\left(p-1\right)/2$ is odd iff $p\equiv3\textrm{ mod }4$ and is even iff $p\equiv1\textrm{ mod }4$. Let $$A\left(N\right)=\sum_{p\leq N}\left(-1\right)^{\left(p-1\right)/2}=\left|\left\{ p\leq N:\, p\equiv1\textrm{ mod }4\right\} \right|-\left|\left\{ p\leq N:\, p\equiv3\textrm{ mod }4\right\} \right|=\pi\left(N;4,1\right)-\pi\left(N;4,3\right).$$ By partial summation $$\sum_{p\leq N}\frac{\left(-1\right)^{\left(p-1\right)/2}}{p}=\frac{A\left(N\right)}{N}+\int_{3}^{N}\frac{A\left(t\right)}{t^{2}}dt$$ then observe that, by prime number theorem on arithmetic progression (using the fact that $4<\left(\log\left(N\right)\right)^{C}$ for some $C>0$) $$\pi\left(N;4,1\right)-\pi\left(N;4,3\right)=O\left(\frac{N}{\phi(4)\log\left(N\right)^{2}}\right)$$ where $\phi(.)$ is the Euler totient function, so $$\sum_{p\leq N}\frac{\left(-1\right)^{\left(p-1\right)/2}}{p}=O\left(\frac{1}{\log\left(N\right)^{2}}+\int_{3}^{N}\frac{1}{t\log\left(t\right)^{2}}dt\right)=O\left(\frac{1}{\log\left(N\right)^{2}}+\frac{1}{\log\left(N\right)}+ \frac{1}{\log(3)}\right)=O\left(1\right).$$ Now take $N\longrightarrow\infty.$

Marco Cantarini
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    How do we know $\pi(N;4,1)-\pi(N;4,3)=O(N/\log^2N)$? Dirichlet on AP's only gives $O(N)$... – Bart Michels Mar 14 '15 at 14:48
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    @barto Note that if we have $\pi\left(N;q,a\right)$ with $q<\log(N)^{C}$ for some $C>0$, holds $$\pi\left(N;q,a\right)=\frac{N}{\phi\left(q\right)\log N}\left(1+O\left(\frac{1}{\log\left(N\right)}\right)\right).$$ – Marco Cantarini Mar 14 '15 at 14:52
  • Oh, I didn't know that. Could you give a reference? – Bart Michels Mar 14 '15 at 14:56
  • @barto Page 7, display (11.9.2) here: http://www.dms.umontreal.ca/~andrew/Courses/Chapter11.pdf – Potato Mar 14 '15 at 14:56
  • Perhaps I am missing something but it seems you show that it is $O(1)$. How does this show convergence? I mean $\sum_{n\le N}(-1)^n$ is also $O(1)$. (I think you method is good, but there might be a small change to be made.) – quid Mar 16 '15 at 11:12
  • @quid The fact that the series is $O(1)$ implies that is convergent. In other therms, we have $$\sum_{p\leq N}\frac{\left(-1\right)^{\left(p-1\right)/2}}{p}<C$$ for some $C>0$ and $C$ is indipendent by $N$. So if you take the limit $N\rightarrow\infty$ we have$$\sum_{p}\frac{\left(-1\right)^{\left(p-1\right)/2}}{p}<C$$ so the series is convergent. – Marco Cantarini Mar 16 '15 at 11:22
  • This is true for sequences with positive terms but not in general. Please have a look at the example I gave, I think it is clearly bounded and clearly not convergent, so there must be some additional consideration in your case. – quid Mar 16 '15 at 11:29
  • Sorry but I don't see the point. The series is convergent (see http://mathworld.wolfram.com/PrimeSums.html) and why it doesn't work in my case? Yes, there is an error in my previous comment, because I should write $$\left|\sum_{p}\frac{\left(-1\right)^{\left(p-1\right)/2}}{p}\right|<C.$$ The alternating series $A(N)$ is $O(N/\log(N)^{2})$, which tends to infinity when $N \rightarrow \infty$. – Marco Cantarini Mar 16 '15 at 11:38