We will denote the set of prime numbers with $\mathcal P$.

We know that the sum $\sum_{n=1}^{\infty}\frac1n$ and $\sum_{n=2}^{\infty}\frac1{n\ln n}$ diverges. It is also known that $\sum_{p \in \mathcal P} \frac1p$ is also diverges, where the sum runs over the $p$ primes.

How could we decide that $\sum_{p \in \mathcal P} \frac1{p\ln p}$ is converges or not?

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    $\begingroup$ Is Mertens' second theorem fair game? $\endgroup$ – Daniel Fischer Nov 25 '14 at 19:32
  • $\begingroup$ @DanielFischer Feel free to post your answer, but Merstens' second theorem is related to prime number theorem quity closely. $\endgroup$ – user153012 Nov 25 '14 at 20:04
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    $\begingroup$ Sure, but Mertens proved it some 22 years before the first proof of the PNT. $\endgroup$ – Daniel Fischer Nov 25 '14 at 20:06
  • $\begingroup$ I decided to use (a weaker version of) the first theorem instead. That way, I could include a complete proof of the used things. [Well, I assume the formula for the multiplicity of $p$ in $n!$ as known.] $\endgroup$ – Daniel Fischer Nov 25 '14 at 20:47
  • $\begingroup$ @user153012 if you disallow anything close to the PNT you'll probably not get any form of answer, you need something approximately that strong to get your result, since it is inherently tied to how quickly primes grow. The PNT gives the exact asymptotic, but anything too much weaker won't be able to give you enough for convergence. $\endgroup$ – Adam Hughes Nov 25 '14 at 22:29

The PNT says that $p_n=n\log n+o(n\log n)$. From this we see by limit comparison that

$$\sum_p{1\over p\log p}$$

converges or diverges depending on as

$$\sum_n{1\over n\log^2 n}$$

which converges by integral test.

  • $\begingroup$ Here $o$ is the little-o notation, right? $\endgroup$ – user153012 Nov 19 '14 at 19:46
  • $\begingroup$ @user153012 Yes. $\endgroup$ – Adam Hughes Nov 19 '14 at 19:53
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    $\begingroup$ I'm really curious where to converges this series, but as I see the convergence is very slow. For the $1000000$th partial sum there isn't any significant digit. $\endgroup$ – user153012 Nov 19 '14 at 20:01

We begin by proving a weaker version of Mertens' first theorem:

$$\left\lvert\sum_{p\leqslant n} \frac{\ln p}{p} - \ln n\right\rvert \leqslant 2\tag{1}$$

for all $n\geqslant 2$.

Although Mertens' first theorem isn't too hard to prove, a complete proof would be too long for this answer, so we only prove

$$\sum_{p\leqslant n} \frac{\ln p}{p} \leqslant 2\ln n\tag{2}$$

for $n\in \mathbb{N}\setminus\{0\}$, which suffices.

For each prime $p\leqslant n$, there are $k(p,n) := \left\lfloor\frac{n}{p}\right\rfloor$ multiples of $p$ that are $\leqslant n$, and hence

$$\prod_{p\leqslant n} p^{k(p,n)} \mid n!$$


$$\sum_{p\leqslant n} \left\lfloor \frac{n}{p}\right\rfloor\ln p \leqslant \ln n!$$

for all $n\geqslant 1$. Thus we have

\begin{align} \sum_{p\leqslant n} \frac{\ln p}{p} &= \frac{1}{n}\sum_{p\leqslant n} \frac{n}{p}\ln p\\ &< \frac{1}{n}\sum_{p\leqslant n} \left(\left\lfloor \frac{n}{p}\right\rfloor + 1\right)\ln p\\ &\leqslant \frac{1}{n}\ln n! + \frac{1}{n}\sum_{p\leqslant n}\ln p\\ &\leqslant \frac{1}{n}\ln (n^n) + \ln n\\ &= 2\ln n. \end{align}

Using $(2)$, we obtain the estimate

$$\sum_{n < p \leqslant n^2} \frac{1}{p\ln p} = \sum_{n < p \leqslant n^2} \frac{\ln p}{p(\ln p)^2} < \frac{1}{(\ln n)^2}\sum_{n < p \leqslant n^2}\frac{\ln p}{p} \leqslant \frac{2\ln (n^2)}{(\ln n)^2} = \frac{4}{\ln n}$$

for every $n \geqslant 2$. Then

\begin{align} \sum_{p} \frac{1}{p\ln p} &= \frac{1}{2\ln 2} + \sum_{k=0}^\infty \sum_{2^{2^k} < p \leqslant 2^{2^{k+1}}} \frac{1}{p\ln p}\\ &\leqslant \frac{1}{2\ln 2} + \sum_{k=0}^\infty \frac{4}{\ln 2^{2^k}}\\ &= \frac{1}{2\ln 2} + \frac{4}{\ln 2} \sum_{k=0}^\infty \frac{1}{2^k}\\ &= \frac{1}{2\ln 2} + \frac{8}{\ln 2}\\ &< +\infty. \end{align}

The obtained bound for the sum is of course ridiculously large, but we were only interested in proving convergence.

  • $\begingroup$ Really nice solution. And it uses a the first theorem of Mertens and not the second, and even a weaker version, so it is quite far from Prime Number Theorem. $\endgroup$ – user153012 Nov 26 '14 at 14:02
  • $\begingroup$ Could you give a reference to a proof of Mertens' first theorem? $\endgroup$ – user153012 Nov 26 '14 at 20:41
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    $\begingroup$ It's theorem 425 (p. 348) in Hardy/Wright, but without explicit bounds on the difference. Mertens' original proof is in "Ein Beitrag zur analytischen Zahlentheorie", Journal für die reine und angewandte Mathematik 78 (1874) pp.46-62, there's a link to a scanned PDF from the University of Göttingen on the wikipedia page (first reference). If you can't read German, it will be hard to follow the first proof, but probably not impossible. $\endgroup$ – Daniel Fischer Nov 27 '14 at 11:37
  • $\begingroup$ @user153012 There are seven multiples of $2$ not exceeding $15$: $2,4,6,8,10,12,14$. Thus we know $2^7 \mid 15!$ (that's not the highest power of $2$ dividing $15!$, but that's enough for our purposes). There's no reason why $k(p,n)$ should be a multiple of $p$. $\endgroup$ – Daniel Fischer May 6 '15 at 21:15
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    $\begingroup$ I didn't take it directly from anywhere, but of course most if not all of the ideas came from stuff I've read somewhere. How big my own contribution is, I cannot tell. Not very big, though. $\endgroup$ – Daniel Fischer May 6 '15 at 23:41

It's already been noted that convergence follows readily from the Prime Number Theorem (Adam Hughes), or even from less precise estimates such as Mertens (Daniel Fischer; the Chebyshev bound $p_k \gg k \log k$ would suffice too) $-$ but also that convergence is frustratingly slow, with the sum over $p > x$ decaying only as $1/\log x$.

Here's another approach, via the Euler product $$ \sum_{n=1}^\infty \frac1{n^s} =: \zeta(s) = \prod_p \frac1{1-p^{-s}} $$ (the product extending over all primes $p$), which makes it practical to estimate $\sum_p 1 / p \log p$ to high accuracy. The numerical value turns out to be 1.636616323351260868569658...; these days, once one has such a decimal expansion, Google will often find a reference, and here the computation is reported in the arXiv preprint

Richard J. Mathar: Twenty Digits of Some Integrals of the Prime Zeta Function (preprint, 2008), arXiv: 0811.4738

(see Table 2.4 at the bottom of page 4).

Taking logarithms of the Euler product we find $$ \log \zeta(s) = \sum_p -\log (1-p^{-s}) = \sum_p \frac1{p^s} + \sum_p \frac1{2p^{2s}} + \sum_p \frac1{3p^{3s}} + \cdots. $$ We can isolate the first contribution $\sum_p 1/p^s$ by taking a suitable linear combination of $\log \zeta(s)$, $\log \zeta(2s)$, $\log \zeta(3s)$, etc., finding $$ \sum_p \frac1{p^s} = \sum_{m=1}^\infty \frac{\mu(m)}{m} \log \zeta(ms), $$ where $\mu$ is the Möbius function.

Now since $1 / \log p = \int_1^\infty p^{-s} ds$, we can integrate the formula for $\sum_p \frac1{p^s}$ to find $$ \sum_p \frac1{p \log p} = \sum_{m=1}^\infty \frac{\mu(m)}{m} \int_1^\infty \log \zeta(ms) \, ds = \sum_{m=1}^\infty \frac{\mu(m)}{m^2} \int_m^\infty \log \zeta(s) \, ds. $$ This clearly converges, because $\int_m^\infty \log \zeta(ms)$ decays as $2^{-m}$ for large $m$, while for $s=1+\epsilon$ we know that $\zeta(s)$ grows as $1/\epsilon$ so $\log\zeta(s)$ grows only as $\log(1/\epsilon)$ which is integrable.

Moreover, we can evaluate $\zeta(s)$ (and $(s-1)\zeta(s)$ near $s=1$) to high precision using Euler-Maclaurin summation of $\sum_{n=1}^\infty 1/n^s$. This makes the formula amenable to known techniques for numerical integration. One such technique is implemented in gp, and the command


takes only a few seconds to return the numerical value reported above.

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    $\begingroup$ Pointing out a few typos, leaving you the typographical choice for the fix: $$\int_1^\infty \frac{ds}{p^s} = \frac{1}{p\log p},$$ you forgot the $p$ in the "Now since" line. I'd probably prefer $\frac{1}{p\log p}$ over $1/(p\log p)$. Then you wrote "because $\int_m^\infty \log \zeta(ms)$", should be one of $\int_1^\infty \log \zeta(ms)$ or $\int_m^\infty \log \zeta(s)$. As usual, very informative answer. $\endgroup$ – Daniel Fischer Nov 29 '14 at 21:45
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    $\begingroup$ Nice answer. I've also studied that way while I've read about Prime zeta function. I have a feeling that this approach is closer to Prime Number Theorem then Daniel Fisher's solution. Thank you for your answer. +1 btw. $\endgroup$ – user153012 Nov 29 '14 at 21:49
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    $\begingroup$ Thanks to both Daniel Fischer and user153012. I'll correct the typos later today, and add a reference to that Wikipedia page on the "prime zeta function". This analysis uses $\zeta(s)$, but only for real $s>1$, which was already known to Euler; the Prime Number Theorem requires much subtler properties of $\zeta(s)$ for complex $s$ outside the half-plane ${\rm Re}(s) > 1$ where $\sum_{n=1}^\infty n^{-s}$ converges. $\endgroup$ – Noam D. Elkies Nov 30 '14 at 20:13

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