We will denote the set of prime numbers with $\mathcal P$.
We know that the sum $$\sum_{n=1}^{\infty}\frac1n \hspace{3mm} \text{and} \hspace{3mm} \sum_{n=2}^{\infty}\frac1{n\ln n}$$ diverges.
It is also known that $$\sum_{p \in \mathcal P} \frac1p$$ also diverges, where the sum runs over the $p$ primes.
How could we decide whether $$\sum_{p \in \mathcal P} \frac1{p\ln p}$$ converges or not?
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.
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.
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.
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.
