is divergent, so for all , if is enough big,
, ie .
Let . By the Cauchy criterion for convergence, there exists an integer N such that whenever n>m>N. In particular, for all p>0. If the sequence is decreasing, then , and therefore . But if p>N then . It follows that if n>2N then . Therefore as .
But we're interested in what happens to . So it would be more useful to know something about rather than . That's the reason for multiplying both sides of the previous inequality by , getting
I hope that makes it a bit clearer. As I said in my previous comment, this proof is fairly heavy-duty analysis. As far as I know, there are no easy shortcuts for proving this result.