Is there a monotonic $f$ such that $\sum f(n)$ diverges but $\sum f(p)$ converges?
Since the $n$'th prime $p_n \approx n \log n$, we should have $p_n \log p_n \approx n (\log n)^2$, so $\sum_p 1/(p \log p)$ should converge, but $\sum_{n \ge 2} 1/(n \log n)$ diverges.
If you let $f(n)=\frac{1}{p_n}$, where $p_n$ is the $n$th prime, the sum on the reciprocals of the primes: $$ \sum_{k=1}^\infty{\frac{1}{p_k}} $$ Diverges, but for $f(p_n)$ the sum over the reciprocal of the super-primes: $$ \sum_{k=1}^\infty{\frac{1}{p_{p_k}}}=.958... $$ Converges, as well as the function being strictly decreasing.