TITLE: $\sum_{p \in \mathcal P} \frac1{p\ln p}$ converges or diverges?
QUESTION [18 upvotes]: 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?

REPLY [2 votes]: 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.