$\sum_{p \in \mathcal P} \frac1{p\ln p}$ 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?
 A: 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.
A: 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.
A: 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.
A: 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.
