Elementary proof that $\sum_n 1/(p_n \log p_n)$ converges for primes $p_n$ 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?
 A: 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.
A: We need not use the prime number theorem to obtain $c_1n\log n <p_n<c_2n\log n$. There are quite elementary arguments for this, see e.g. How prove that $P_{2n}<(n+1)^2$.
Then your argument should show that the series $\sum_n \frac{1}{p_n\log n}$ converges.
A: Let $k = p_n, k > 2$ and $n = \pi(k).$ Then 
$$n = \pi(k) < \frac{6k}{\log k}= \frac{6p_n}{\log p_n}  $$
so $$p_n > \frac{1}{6}n \log p_n > \frac{1}{6}n \log n.\tag1$$
This relies on another fact from Apostol, that 
$\pi(n) < \frac{6n}{\log n}.$
Now the inequality (1) gets you the comparison in the OP. Proof of the line above is given at p. 83 of Apostol and does not rely on the prime number theorem. I wouldn't call it 'easy.'
