So let's look at n^m. The condition is sum_n n^-m ; and that diverges for m smaller or equal 1. If you have a sequence that grows asymptotically like n^m it me

So let's look at n^m. The condition is sum_n n^-m ; and that diverges for m smaller or equal 1. If you have a sequence that grows asymptotically like n^m it means that f(n)/n^m goes to a constant asymptotically. So the question is if this implies that sum 1/f(n) converges. I feel intuitively that it should be possible to prove that. A sketch: For every epsilon there is an N such that |1/f(n) - 1/C n^-m| < epsilon n^-m for n > N. So if we take the difference between the reciprocal partial sums, then that is bounded by eps * the partial sum. That is finite exactly if the reciprocal sums converge. Thus the convergence behaviour is the same for this case...