TITLE: Elementary proof that $\sum_n 1/(p_n \log p_n)$ converges for primes $p_n$
QUESTION [5 upvotes]: The prime number theorem says that the $n$th prime number is $p_n = \Theta(n \log n)$, so the series $\sum_n 1/(p_n \log p_n)$ should converge by comparison to $\sum_n 1/n (\log n)^2$. However this seems like overkill using deep mathematics. Is there a more elementary proof that $\sum_n 1/(p_n \log p_n)$ converges?

REPLY [4 votes]: A combinatorial proof of the bounds $$\frac{x}{\log x} \ll \pi(x)\ll \frac{x}{\log x}$$ can be found in the following two Math.StackExchange answers:  
How to prove Chebyshev's result: $\sum_{p\leq n} \frac{\log p}{p} \sim\log n $ as $n\to\infty$? and Are there any Combinatoric proofs of Bertrand's postulate? Also see this answer for an application: Chebyshev: Proof $\prod \limits_{p \leq 2k}{\;} p > 2^k$ 
The bound is obtained by looking at the central binomial coefficient $\binom{2n}{n}$, and the primes dividing it. This approach is due to Erdos. It is not hard to see that the convergence of your series follows from these bounds.