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Let $p_i$ denote the $i^{\rm th}$ prime number, and let $\lfloor x \rfloor$ denote the greatest integer smaller than or equal to $x$. Then does the sum defined by:

$$ \sum_{n=0}^{\infty} \frac{1}{p_{\lfloor\pi n\rfloor}} $$

converge?

Computing the partial sum up to $n=3\times10^7$ yields a value of about $\approx 0.890694\ldots$, and Wolfram Alpha gives the following plot, suggesting that the sum might be convergent:

$\hspace{6cm}$enter image description here

Note that Wolfram does not explicitly state that the sum is non-convergent, as it does for instance in the case $\sum_{n=0}^{\infty} \frac{1}{p_{2n}}$.

Klangen
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    It surely diverges. Matijasevic said, the sum over all known primes $p$ of $1/p$ is less than 5 – and it always will be. The divergence may be very slow, but it is certain. – Gerry Myerson Oct 12 '17 at 10:13
  • But we know that sums of reciprocals of certain subsets of primes can converge, see https://math.stackexchange.com/questions/1875520/does-the-sum-of-reciprocals-of-this-subset-of-the-primes-converge and https://math.stackexchange.com/questions/912852/define-an-infinite-subset-of-primes-such-that-the-sum-of-reciprocals-converges – Klangen Oct 12 '17 at 10:15
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    Sure, but as you say, $\sum p_{2n}^{-1}$ diverges, and so do $\sum p_{3n}^{-1}$ and $\sum p_{4n}^{-1}$ and so on, and your sum is in that family. – Gerry Myerson Oct 12 '17 at 10:22
  • @GerryMyerson I was thinking the same thing. What intrigued me is that Wolfram does not say they diverge, while for $\sum p_{2n}^{-1}$ it does – Klangen Oct 12 '17 at 10:26
  • I don't know what Wolfram says, since you haven't told us where to find it. – Gerry Myerson Oct 12 '17 at 10:35
  • @GerryMyerson Simply put sum 1/Prime[floor(pi*n)], for n=1 to infinity into https://www.wolframalpha.com – Klangen Oct 12 '17 at 10:39
  • And if you put the same for $2n$ into alpha, it explicitly tells you the sum diverges? I guess alpha is clever, but not quite clever enough. – Gerry Myerson Oct 12 '17 at 10:53
  • @GerryMyerson Yes, it says "sum does not converge": https://www.wolframalpha.com/input/?i=sum+1%2FPrime%5B2*n%5D,+for+n%3D1+to+infinity – Klangen Oct 12 '17 at 10:56

1 Answers1

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We have $\lfloor \pi n \rfloor \leq 4n$. It is therefore sufficient to show the divergence of $$\sum_{n=0}^{\infty} \frac{1}{p_{4n}}$$ Assume it does converge. Note that $$\frac{1}{p_{4n+k}} \leq \frac{1}{p_{4n}}$$ for all $k\in\mathbb{N}$. Thus, the sum $$\sum_{n=0}^{\infty} \frac{1}{p_{4n+k}}$$ converges for all $k \in \mathbb{N}$. Hence $$\infty > \sum_{k=0}^3 \sum_{n=0}^{\infty}\frac{1}{p_{4n+k}}=\sum_{n=0}^{\infty} \sum_{k=0}^{3}\frac{1}{p_{4n+k}}=\sum_{n=0}^{\infty} \frac{1}{p_n}$$ But the right hand sum diverges.

This is an instance of a more general fact which is not related to prime numbers: If a sum $$\sum_{n=0}^{\infty} a_n$$ with $a_0 \geq a_1 \geq .. $ an all $a_n \geq 0$ diverges, then for each $k \in \mathbb{N}$, the sum $$\sum_{n=0}^{\infty} a_{kn}$$ diverges. The proof is the same as above.

russoo
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  • I disagree with the first line in your answer. The sequence $p_{\lfloor\pi n\rfloor}$ starts with: $5,13,23,37,47$, while $p_{4n}$ starts with: $7,19,37,53,71$. The sum of their reciprocals cannot therefore be compared. – Klangen Oct 12 '17 at 11:00
  • Maybe I am confused. But, as you write, $p_{\lfloor \pi n\rfloor} \leq p_{4n}$, therefore $(p_{4n})^{-1} \leq (p_{\lfloor \pi n\rfloor})^{-1}$. – russoo Oct 12 '17 at 11:06
  • I agree with that. However, I do not with the statement "It is therefore sufficient to show the divergence of $\sum p_{4n}^{-1}$", since we are talking about indices of prime numbers, not the prime numbers themselves. Therefore the sequences are completely different, and the reciprocals also. – Klangen Oct 12 '17 at 11:12
  • But we have $\sum p_{4n}^{-1} \leq \sum p_{\lfloor \pi n \rfloor}^{-1}$. Divergence follows then by the comparision test for series with non-negative summands. – russoo Oct 12 '17 at 11:14
  • Did you mean $\sum p_{4n}^{-1} \geq \sum p_{\lfloor \pi n\rfloor}^{-1}$? Also, how can you compare two series with completely different terms? It might be that all the terms of one are smaller than all the terms of the other, but this does not mean that if one diverges, so does the other... – Klangen Oct 12 '17 at 11:28
  • As I worte above, $p_{4n}^{-1} \leq p_{\lfloor \pi n\rfloor}^{-1}$. And this holds for all(!) $n$. This implies $\sum_{n=0}^{\infty} p_{4n}^{-1} \leq \sum_{n=0}^{\infty} p_{\lfloor \pi n \rfloor}^{-1}$. – russoo Oct 12 '17 at 11:32
  • And: For two series $\sum_{n=0}^{\infty} a_n$, $\sum_{n=0}^{\infty} b_n$ with $0 \leq a_n \leq b_n$ for all $n$, we have $\sum_{n=0}^{\infty} a_n \leq \sum_{n=0}^{\infty} b_n$. Thus divergence of $\sum_{n=0}^{\infty} a_n$ implies divergence of $\sum_{n=0}^{\infty} b_n$. (That is the comparison test!) To understand my argument from above, set $a_n=p_{4n}^{-1}$ and $b_n=p_{\lfloor \pi n \rfloor}^{-1}$. – russoo Oct 12 '17 at 11:44
  • If that were true all series would diverge, because you can always find a $b_n$ such that $\sum_{n=0}^{\infty} a_n \leq \sum_{n=0}^{\infty} b_n$ – Klangen Oct 12 '17 at 11:59
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    The last assertion in the answer is wrong, an assumption like monotonicity of the sequence $(a_n)$ is required. Otherwise, consider $a_n = n^{-2 - (-1)^n}$, then $\sum a_{kn}$ converges for even $k$. – Daniel Fischer Oct 12 '17 at 12:12
  • @Pickle You misunderstand. The sequence $(b_n)$ is given, and its convergence or divergence shall be determined. If you can find a sequence $(a_n)$ such that for all $n$ you have $0 \leqslant a_n \leqslant b_n$, and such that $\sum a_n$ diverges, then you conclude that $\sum b_n$ diverges too. In the situation of this question, $b_n = p_{\lfloor \pi n\rfloor}$, and $a_n = p_{4n}$. – Daniel Fischer Oct 12 '17 at 12:16
  • @DanielFischer: Thanks for your remark! I edit my answer. Also, your counterexample is quite nice. – russoo Oct 12 '17 at 12:19
  • even though your argument still holds for $p_{3n}^{-1}$ that is diverge sum, actually and $p_{c n}^{-1}$ is diverge sum given that $c>0$ and constant. – Ahmad Oct 13 '17 at 15:47
  • easier to prove by PNT, which imply that $p_n > n \ln n$ so $\frac{1}{\pi n \ln(\pi n)}$ is obviously diverges. – Ahmad Oct 13 '17 at 15:49