I am studying analytic number theory at the time and I came across with this. I was wondering if there is a somehow smart way to prove the following without using prime number theorem(but probably use some results for approximation of the primes by $n\log n$ or something like that, maybe approximation of $π(x)$ by $x/\log x$ or something like these results which are usually proven just before the prime number theorem) and without any summation formula:
Let $a(n)$ be a nonincreasing sequence of positive numbers. Show that $\sum a(p)$ -sum over all primes- converges if and only if $\sum_{n=2}^{\infty} a(n)/\log n$ converges.
I have seen 2 questions here about this where they use prime number theorem and summation formulas, but I have a strong feeling that it can be proven without these, I have tried it but nothing seems to work out well. My idea is to work somehow like proving the Cauchy condensation theorem for series and maybe use somehow the fact that there are constants $a$, $b$ positive such that $an\log n<p(n)<bn\log n$ for all $n$ from some $N$ and after -$p(n)$ is the $n$th prime. Or maybe the fact that $(1/6)n/\log n<π(n)<6n/\log n$ for all $n=2,3,...$, or maybe something like that. Any help or ideas whould be much appreciated.