if a1+a2+a3.. decreasing sequence of positive numbers
How to prove that if a1+a2+a3 ... converges then lim (if n to infinity) of na(subn)=0
I think you need to roll up your sleeves and wheel out some heavy-duty epsilons in order to prove this.
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 .
In the sum , each term is greater than or equal to the last term (since the sequence is decreasing). So the sum is That gives the inequality
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.