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I was wondering if the series

$$\sum\limits_{n=1}^\infty\left(\frac n{p_n}\right)^k$$

converges for $k>1$, where $p_n$ is the $n^\textrm{th}$ prime number?

Edit: Thinking about this a bit more, it is known that $p_n$ is bounded by $p_n<n(ln(n)+ln(ln(n)))$ for big enough $n$, meaning that

$$\left(\frac n{p_n}\right)^k > \left(\frac 1{ln(n)+ln(ln(n))}\right)^k $$

for these $n$ and since

$$\sum\limits_{n=2}^\infty\frac 1{nln(n)}$$ diverges, does this imply the divergence of $$\sum\limits_{n=1}^\infty\left(\frac n{p_n}\right)^k$$ for any $k$?

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  • $\begingroup$ In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. $\endgroup$ – Sahiba Arora Jul 22 '17 at 18:17
  • $\begingroup$ I did not really encounter this problem anywhere but was just thinking about series that might be interesting and could not find anything to this one, so Im just asking out of curiosity. I know the proof showing that $\sum\limits_{n=1}^\infty (\frac 1{p_n})^k$ diverges for $k=1$ (but of course converges for $k>1$). $\endgroup$ – Bertrand Jul 22 '17 at 18:34
  • $\begingroup$ Maybe you can use the prime number theorem. $\endgroup$ – mathreadler Jul 22 '17 at 18:35
  • $\begingroup$ What is $p_n$ ? $\endgroup$ – Sahiba Arora Jul 22 '17 at 18:36
  • $\begingroup$ sorry, $p_n$ is the n-th prime number. $\endgroup$ – Bertrand Jul 22 '17 at 18:36
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Since $p_n \approx n \ln n$ and $\ln(x) \lt x^c$ for any $c > 0$ and large enough $x$,

$\begin{array}\\ \sum\limits_{n=1}^\infty\left(\frac n{p_n}\right)^k &\approx \sum\limits_{n=1}^\infty\left(\frac n{n\ln n}\right)^k\\ &= \sum\limits_{n=1}^\infty\left(\frac1{\ln n}\right)^k\\ &\gt \sum\limits_{n=1}^\infty\left(\frac1{n^c}\right)^k \qquad\text{for any } c > 0\\ &= \sum\limits_{n=1}^\infty\frac1{n^{ck}}\\ &= \sum\limits_{n=1}^\infty\frac1{n^{1/2}} \qquad\text{choose } c = \frac1{2k}\\ \text{Diverges}\\ \end{array} $

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  • $\begingroup$ It is an elementary consequence of the Prime Number Theorem that $\lim_{n\to \infty}(n\ln n)/p_n=1.$(It does take some work.) So $n/p_n>1/(2\ln n)$ for all but finitely many $ n.$ $\endgroup$ – DanielWainfleet Jul 23 '17 at 4:16

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