The sum of the reciprocal of primeth primes

A few days ago, a friend of mine taught me that the sum of the reciprocal of primeth primes $$\frac{1}{3}+\frac{1}{5}+\frac{1}{11}+\frac{1}{17}+\frac{1}{31}+\cdots$$ converges.

Does anyone know some papers which have a rigorous proof with the convergence value?

• There are only about $20$ primeth primes that have been calculated (see OEIS A007097). Obviously, this sum is smaller than the sum of reciprocals of power of $2$, so it would quickly converge. EDIT: Sorry, that's a different sequence (which you might find interesting regardless, so I'll leave this comment here for a few more minutes). – barak manos Feb 16 '15 at 18:39
• @barakmanos: You're talking about a different sequence than the author. Yes, the sum of the reciprocals of A007097 converge very quickly, but so do the sum of the reciprocals of A006450 (albeit much more slowly). – Charles Feb 16 '15 at 18:40
• @Charles: Yes, I just noticed (and noted) that. – barak manos Feb 16 '15 at 18:41
• Is the convergence value even known? – RghtHndSd Feb 16 '15 at 20:27
• – nathan.j.mcdougall Sep 2 '15 at 5:54

It is a duplicate. We have $p_n\gg n\log n$ by Chebyshev's theorem (or the PNT) hence $$p_{p_n}\gg p_n \log n \gg n\log^2 n$$ and $$\sum_{n\geq 1}\frac{1}{n\log^2 n}$$ is convergent by Cauchy's condensation test.
• - or the integral test, since $\int 1/(t\log^2t)\,dt = -1/\log t+C$ – Greg Martin Feb 16 '15 at 18:41
Jack has already shown that the sequence converges. By summing the primeth primes up to $10^{11}$ and taking an integral to cover the missing terms I estimate that the reciprocal sum is about 1.05. The sum up to $10^9$ is 0.9904, the sum up to $10^{10}$ is 0.9960, and the sum up to $10^{11}$ is 1.0005.