# Convergent series

1. May 28, 2005

### steven187

hello all

iv been workin on this problem its kind of awkward check it out

{an} is a decreasing sequance, an>=0 and there is a convergent series Sn with terms an
we need to prove that the limit of nan is 0

i first started of a sequence bn=an+1+an+2+.......+a2n
then I showed that the limit of bn is also 0 where do i go from here

Last edited: May 28, 2005
2. May 28, 2005

### mathwonk

use the hypothesis, that they are decreasing. this immediately implies your result for even n, and then also the result for all n.

[how did you think of that way of beginning? was it a hint? i.e. the first step was the clever part.]

3. May 29, 2005

### steven187

well in terms of how i began i was trying to think of a sequence that converges to zero so that i can bound the sequence nan from below and i was trying to look for something else to bound it by from above so that i can use either the comparision test or the squeze theorem but no matter how much i play with it i cant find anything to bound it by from above so far this is what iv done

bn=an+1+an+2+.......+a2n then i showed
lim n->infinity bn= 0
bn=an+1+an+2+.......+a2n
<=an+an+.......+an
=nan this is where i get confused I obviously cant bound it by the sequence an because nan>=an so where do i go from here

by the way how do you use this latex stuff im sure it would be pretty useful and quik

4. May 29, 2005

### mathwonk

what you have shown, proves the result for even n, assuming decreasingness. then it also follows for odd n, again using decreasingness.

i.e. 2n a(n) goes to zero if and only if n a(n) does.