Show that $$\sum_{k=1}^\infty (p_k\log(p_k))^{-1} \mbox{ converges.} $$ Here $p_k$ is the $k$th prime number.
This is in Elementary number theory. I tried doing the integral test but I was not given bounds on the summation so I did not know the bounds on the integral.
Rosser proved that the $k$th prime is greater than $k\log k$, and Dusart proved even more: $$ p_k > k(\log k + \log\log k -1) \qquad\mbox{ for }k\ge 2. $$ Therefore each partial sum of your series $$ \sum\limits_{k=1}^\infty {1\over p_k \log p_k} $$ is bounded by the respective partial sum of $$ \sum\limits_{k=1}^\infty {1\over k\log k \log p_k} $$ which is in turn bounded by $$ \sum\limits_{k=1}^\infty {1\over k\log^2 k}. $$ The latter series converges.