4
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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). $\endgroup$ – barak manos Feb 16 '15 at 18:39
  • 2
    $\begingroup$ @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). $\endgroup$ – Charles Feb 16 '15 at 18:40
  • $\begingroup$ @Charles: Yes, I just noticed (and noted) that. $\endgroup$ – barak manos Feb 16 '15 at 18:41
  • $\begingroup$ Is the convergence value even known? $\endgroup$ – RghtHndSd Feb 16 '15 at 20:27
  • 3
3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ - or the integral test, since $\int 1/(t\log^2t)\,dt = -1/\log t+C$ $\endgroup$ – Greg Martin Feb 16 '15 at 18:41
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.