I use [1] (in spanish) for the sequence of positive terms defined by

$$ a_k = \begin{cases} \frac{1}{k}(\frac{1}{p_k}+\frac{1}{p_k+2}), & \text{for the kth twin prime pair} \\ 0, & \text{if there are finitely many twin primes pairs and $k>$ last of them} \end{cases}$$

Then since (as in the last remark is claimed, the first condition, this is the convergence of $\sum_{k=1}^{\infty}a_k$ follows from the convergence of second series $\sum_{k=1}^{\infty}\sum_{j=k}^{\infty}a_j$) we have the claimed result, by Brun's theorem and the first statement in the proof:

Since $$\sum_{j=1}^{\infty}ja_j=\sum_{\text{twin primes}}(\frac{1}{p_j}+\frac{1}{p_j+2})<\text{Brun's constant}$$ is convergent, then for $p>1$, using Hölder's theorem and Stoltz's theorem, I ask if with the last remark that claims the statement also for $p\geq 1$, then $p=1$, the following questions is answered as true

Question. Can you sure that, following this way for the sequence of twin primes, $$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n k\sum_{j=k}^{2k-1}a_j=0?$$ Or have I claimed a misunderstanding? I say when take the cited sequence of strictly positive numbers or taking $p=1$. Truly it is possible find a infinite sequence of primes (any sequence) satisfying $$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n k\sum_{j=k}^{2k-1}a_j=0,$$ for $a_j=\frac{1}{j}(\frac{1}{\mathcal{P}_j}+\frac{1}{\mathcal{P}_j+2})$, $j\geq 1$? Give an heuristic or a proof. Or is dificult, impossible or false?

Thanks in advance.


The Problem section is always free in this journal, and some exercises are easily readable (more than in english!)

[1] La GACETA de la Real Sociedad Matemática Española, VOL. 12, n.º 4 Año 2009, PROBLEMAS Y SOLUCIONES, page 700 PROBLEMA 116 http://gaceta.rsme.es/abrir.php?id=896


Using standard results $a_k \ll \frac{1}{k^2\log^2k}$ and so $$ k\sum_{j=k}^{2k-1}a_j \ll \frac{1}{\log^2k} $$ and $$\frac{1}{n}\sum_{k=2}^n k\sum_{j=k}^{2k-1}a_j \ll \frac{1}{n}\sum_{k=2}^n\frac{1}{\log^2k} \le \frac{1}{n}\int_{x=2}^{n+1}\frac{dx}{\log^2x} \ll \frac{1}{\log^2n}$$ (where the first term has been dropped only to avoid issues with $\log1$ in the upper bound) and so $$ \lim_{n\to\infty}\frac{1}{n}\sum_{k=2}^n k\sum_{j=k}^{2k-1}a_j=0 $$ which obviously yields $$ \lim_{n\to\infty}\frac{1}{n}\sum_{k=2}^n k\sum_{j=k}^{2k-1}a_j=0 $$ since $a_1=8/15$ changes the sum by only $O(1/n).$

  • $\begingroup$ I take notes from your answer, it incredible, very thanks much. $\endgroup$ – user243301 Jan 1 '16 at 10:25
  • $\begingroup$ @JuanLG: Note that the first result can be derived with the Brun sieve (with a bad constant which I suppressed with the Vinogradov notation $\ll$). $\endgroup$ – Charles Jan 5 '16 at 16:54
  • $\begingroup$ Thanks when I will study this sieve I remember it. It would to be useful for another users thanks Charles $\endgroup$ – user243301 Jan 5 '16 at 20:21

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