question about convergences for prime related sums

Question

Let $a_n$ be a nonincreasing sequence of positive numbers.

Show that $\sum \limits_{p \text{ is prime}} a_p$ where $p$ runs over all prime numbers converges if and only if $\sum \limits_{n=2}^{\infty} \frac{a_n}{\log n}$ converges ?

Answer 1

The primes have logarithmic density in the natural numbers, this should give us some intuition as to why this is true.

For the proof, we make use of partial summation, which is the analogue of integration by parts for sums: $$\sum_{n\leqslant N}f(n)\,\Delta g(n) = f(N)g(N+1)-f(1)g(1)-\sum_{n<N}g(n+1)\Delta f(n).$$ Here $\Delta f(n)$ denotes the forward difference $f(n+1)-f(n)$.

For the first sum, take $f(n) = a_n$ and $g(n) = n$. Notice that $\Delta g(n)=1$, and since we're summing over primes, $\sum_{p\leqslant N}g(p) = \sum_{n\leqslant N}\pi(n)$. Thus we have

\begin{align*} \sum_{p\leqslant N}a_p = \sum_pa_p\cdot1 &= a_N\,\pi(N+1) - \sum_{n<N}\pi(n+1)\,(a_{n+1}-a_{n}) \\ &\sim a_N\, \frac{N+1}{\log(N+1)} - \sum_{n<N}\frac{n+1}{\log(n+1)}\,(a_{n+1}-a_{n}) \tag{$*$} \end{align*} by the prime number theorem.

Treating the second sum with the same $f$ and $g$, we get $$\sum_{2\leqslant n\leqslant N}\frac{a_n}{\log n} = \sum_{2\leqslant n \leqslant N}\frac{a_n}{\log n}\cdot 1 = \frac{a_N}{\log N}\,(N+1)-\sum_{2\leqslant n < N}n\Big(\frac{a_{n+1}}{\log(n+1)}-\frac{a_{n}}{\log n}\Big),$$ which clearly converges if and only if $(*)$ converges.