$\sum_{p \in \mathcal P} \frac1{p\ln p}$ converges or diverges?

Question

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?

Answer 1

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.

Answer 2

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

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

and

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

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.

Answer 3

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

sum(m=1,199,moebius(m)*intnum(s=m,200,log(zeta(s)))/m^2)

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

Answer 4

There is a theorem of Chebyshev (see p. 384 here) that settles this question by converting it into a question about a series over all integers greater than 1: if $\{a_n\}$ is a sequence of real numbers that is positive and decreasing for all large $n$, then $\sum_{n \geq 1} a_n$ converges if and only if $\sum_{p} a_p(\log p)$ converges, where the second series runs over the primes. The proof relies on the sums $\theta(x) = \sum_{p \leq x} \log p$ being bounded above and below by a constant multiple of $x$: $ax < \theta(x) < bx$ for some positive $a$ and $b$ and all large $x$ (or $x \geq 2$). That is weaker than the Prime Number Theorem.

If you want to know whether $\sum_p 1/(p \log p)$ converges, we want to take $a_p = 1/(p(\log p)^2)$ for prime $p$. So use $a_n = 1/(n(\log n)^2)$ for $n \geq 2$. Since $\sum_{n \geq 2} 1/(n(\log n)^2)$ converges (integral test), the series $\sum_p 1/(p\log p)$ also converges.