Prove that $\sum_{p \in \mathcal P} \frac1{p\ln \ln p}$ is divergent In this question I've asked to decide the convergency of the series $\sum_{p \in \mathcal P} \frac1{p\ln p}$.
Now I ask you to show that the series 
$$\sum_{p \in \mathcal P} \frac1{p\ln \ln p}$$
is divergent, where $\mathcal P$ is the set of all primes.
Maybe there is analogous ways as we've seen the proofs I referred above. I prefer techniques like what was used on the other problem, but of course any proof would by nice.
 A: Recall again the prime number theorem, which gives that
$$ p_n \approx n\log n.$$
Then the sum we want can be compared to
$$ \sum_p \frac{1}{p \log \log p} \sim \sum_n \frac{1}{n\log n \log \log n}.$$
I claim this latter sum diverges based on the integral test for convergence.

Claim: The following series diverges:
  $$\sum_n \frac{1}{n \log n \log \log n} \tag{1}$$

Proof: Notice that $f(x) = \dfrac{1}{x \log x \log \log x}$ is a positive, decreasing, well-defined function for $x > 100$, say. So we may apply the integral test, and the sum in $(1)$ converges if and only if the integral below converges:
$$ \int_{100}^\infty \frac{1}{x \log x \log \log x} dx.$$
Perform the $u$-substitution $u = \log \log x$, and note that $du = \dfrac{dx}{x\log x}$. So this integral behaves like
$$ \int_{\log \log 100}^\infty \frac{1}{u} du = \left.\log u \right\rvert_{\log \log 100}^\infty \approx \log \log \log \infty,$$
which diverges. Note also that this means the integral (and therefore the sum) diverges like $\log \log \log n$, which is so unbelievably slow that it's hard to comprehend. But nonetheless, it diverges. $\diamondsuit$
A: Long story short,
$$ p_n \leq C n \log n $$
by Chebyshev's theorem ($\pi(n)\geq D \frac{n}{\log n}$), hence the series is lower bounded by a multiple of
$$ \sum_{n\geq 3}\frac{1}{n\log n \log\log n}$$
that diverges due to Cauchy condensation test (applied twice).
