How can I prove the following sum converges, where $s>1$ and the sum is over all primes. $$\displaystyle\sum_{p}^{}\displaystyle\sum_{k=1}^{\infty}\frac{\log p}{p^{ks}}$$ I've tried grouping terms in various ways, but it seems difficult to prove, because of being over all primes and all powers. More elementary methods would also be appreciated, without assuming any facts about the Riemann zeta function

  • $\begingroup$ try first with $\sum_{p^k} p^{-ks}$ can you prove it is a sub-series of $\sum_{n=1}^\infty n^{-s} = \zeta(s)$, which is absolutely convergent for $Re(s) > 1$ ? in the same way $\sum_{p^k} \log(p) p^{-ks} = -\zeta'(s)/\zeta(s)$ is dominated term-by-term by $\sum_{n=1}^\infty \log(n) n^{-s} = -\zeta'(s)$, and $\sum_{p^k} \frac{p^{-ks}}{k} = \ln \zeta(s)$ is dominated term by term by $\sum_{n=1}^\infty n^{-s} =\zeta(s)$, everything being absolutely convergent for $Re(s)>1$ $\endgroup$ – reuns Jun 5 '16 at 13:30
  • $\begingroup$ the only fact I'm using is that $\sum_{n=1}^\infty n^{-s}$ and $\sum_{n=1}^\infty \ln(n) n^{-s}$ are absolutely convergent for $Re(s) > 1$... $\endgroup$ – reuns Jun 5 '16 at 13:34
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    $\begingroup$ Why is there a problem? Everything is positive hence the double sum is $$\sum_p\log p\sum_{k\geqslant1}p^{-ks}=\sum_p\frac{\log p}{p^s-1}$$ and, when $p\to\infty$, $$\frac{\log p}{p^s-1}\sim\frac{\log p}{p^s}$$ hence one is left with $$\sum_p\frac{\log p}{p^s}.$$ Now, if $s>1$, ... while if $s=1$, ... $\endgroup$ – Did Jun 5 '16 at 13:35
  • $\begingroup$ @Did : $\sum_p \log(p) p^{-s}$ (as $\sum_{p^k} \frac{\log(p^k)}{k} (p^k)^{-s}$ ) being dominated term-by-term by $\sum_{n=1}^\infty \log(n) n^{-s}$. Note that for proving it diverges on $Re(s) < 1$ and at $s=1$, you'll need to prove that $\sum_{p^k} \log(p) p^{-sk} = -\zeta'(s)/\zeta(s)$ $\endgroup$ – reuns Jun 5 '16 at 13:40
  • $\begingroup$ @user1952009 Indeed, the $n$th prime is at least $n$ and $x\mapsto(\log x)/x^s$ is decreasing for $x$ large enough hence, neglecting a finite number of terms, $\sum\limits_p(\log p)/p^s\leqslant\sum\limits_n(\log n)/n^s$ -- which should yield the first "..." in my comment. $\endgroup$ – Did Jun 5 '16 at 13:42

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