# Rate of convergence of series of squared prime reciprocals

It is well known that $\sum_{p \text{ prime}} \frac{1}{p}$ diverges, and in fact - it behaves like log of the harmonic series: $$\sum_{p \le x} \frac{1}{p} = \log \log x + O(1).$$ It is also well known that $\sum\limits_{p \text{ prime}} \frac{1}{p^2}$ converges. What is known about the rate? Letting $C = \sum\limits_{p \text{ prime}} \frac{1}{p^2}$, what can be said about $C - \sum\limits_{p \le x} \frac{1}{p^2}$?

I am reading an article (a survey of Artin's Primitive Root Conjecture - which follows from the GRH). I am trying to understand what are the condition on functions $f_1\le f_2$ tending to infinity in order that $$\sum_{f_1(x) \le p \le f_2(x)} \frac{1}{p^2} = O\left(\frac{1}{\log x}\right).$$ Of course I can take $f_1(x) = \log\log\log x$, $f_2(x) = \log\log x$, but I want the general conditions.

-
The sum from $x$ to infinity of $1/p^2$ is bounded by the sum from $x$ to infinity of $1/n^2$ which, by comparison with the integral, is on the order of $1/x$. 29% accept rate - do you understand the value of accepting answers to your questions here? –  Gerry Myerson Jan 14 '12 at 23:20

Lets rearrange the sum $C-\sum_{p\leq x}\frac{1}{p^{2}}=\sum_{p>x}\frac{1}{p^{2}}$. Using integration by parts this is $$\sum_{p>x}\frac{1}{p^{2}}=\int_{x}^{\infty}\frac{1}{t^{2}}d\left(\pi(t)\right)=\frac{\pi(t)}{t^{2}}\biggr|_x^\infty+2\int_{x}^{\infty}\frac{\pi(t)}{t^{3}}dt.$$ Using the prime number theorem, that is the asymptotic for $\pi(x),$ you can deduce that the quantity on the right hand side is $\sim\frac{1}{x\log x},$ which is your rate of convergence.
Just worth noting, this is exactly what we would expect. The tail of the sum over all integers has size $\frac{1}{x}$, that is $\sum_{n>x} \frac{1}{n^2}\sim \frac{1}{x}$ and the primes occur with density $\frac{1}{\log n}$ around $n$, so we would expect the tail of the sum to be of size $\frac{1}{x\log x}$. Partial summation/integration allows to prove this.
Edit: Replaced $\asymp$ with $\sim$, since as pointed out by Greg Martin in the comments, the Prime Number Theorem is strong enough to yield this/