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

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.
