Is $\sum_{n=1}^{{\infty}}\frac{1}{P_{3n}}$ convergent?

Question

Is this sum below convergent? ($P_{n}$ is the nth prime.)

$$\sum_{n=1}^{{\infty}}\frac{1}{P_{3n}}$$

Note that $\frac{1}{P_{3n+1}} < \frac{1}{P_{3n}}$ and $\frac{1}{P_{3n+2}} < \frac{1}{P_{3n}}$ so if this series is convergent then so is $\sum \frac{1}{P_{3n+1}}$ and $\sum\frac{1}{P_{3n+2}}$ and therefore $\sum \frac{1}{P_{3n}} + \frac{1}{P_{3n+1}} + \frac{1}{P_{3n+2}} = \sum \frac{1}{P_n}$. Do you know if this series is convergent? In not you can check out this question. — Winther, Dec 30 '15 at 19:59
btw once you have solved this you can try to generalize the problem and try to show that the series $\sum \frac{1}{p_{An+B}}$ diverges for any choice of integers $A,B$. It's not that much more complicated. — Winther, Dec 30 '15 at 20:09
And after that, that $\sum\frac1{p_{\alpha(n)}}$ diverges for any choice of the function $\alpha:\mathbb N\to\mathbb N$ such that $\limsup\alpha(n)/n$ is finite. — Did, Dec 31 '15 at 10:19

score 6 · Answer 1 · answered Dec 30 '15 at 20:03

The series $$\frac{1}{p_1}+\frac{1}{p_2}+\frac{1}{p_3}+\frac{1}{p_4}+\frac{1}{p_5}+\cdots$$ diverges, by a result of Euler. If your series converged, then by Comparison so would $\sum \frac{1}{p_{3n+1}}$ and $\sum \frac{1}{p_{3n+2}}$, and therefore so would Euler's series.

score 0 · Answer 2 · answered Dec 31 '15 at 09:30

Another way is to use the Prime Number Theorem, that tells us $$p_{n}\sim n\log\left(n\right)$$ and so $$\sum_{n\geq1}\frac{1}{p_{3n}}\sim\frac{1}{3}\sum_{n\geq2}\frac{1}{n\left(\log\left(n\right)+\log\left(3\right)\right)}\sim\frac{1}{3}\sum_{n\geq2}\frac{1}{n\log\left(n\right)}$$ and the RHS diverges.

Is $\sum_{n=1}^{{\infty}}\frac{1}{P_{3n}}$ convergent?

2 Answers2