• Support PF! Buy your school textbooks, materials and every day products Here!

If a series converges with decreasing terms, then na_n -> 0

  • #1
1,456
44

Homework Statement


Prove that if ##(a_n)## is a decreasing sequence of positive numbers and ##\sum a_n## converges, then ##\lim na_n = 0##

Homework Equations




The Attempt at a Solution


Let ##\epsilon >0##. By the Cauchy criterion there exists an ##N\in \mathbb{N}## such that ##\forall n\ge m\ge N##, we have that ##|\sum_{k=m+1}^{n}a_k|<\epsilon##. But the sequence is decreasing, so ##|(n-m)a_n|\le |\sum_{k=m+1}^{n}a_k|<\epsilon##. So we have that ##|na_n-ma_n|<\epsilon## for all ##\epsilon>0##. So ##na_n=ma_n## for all ##n\ge m\ge N##. Since the tails of these sequence are the same eventually, they have the same limit. Since ##\sum a_n## converges, we see that ##\lim ma_n = 0##. Hence ##\lim n a_n = 0##.
 

Answers and Replies

  • #2
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,519
734
That argument doesn't work. Given ##\epsilon > 0## there exists ##N## such that...
The problem is that the ##N## depends on ##\epsilon##. So the (obviously false) conclusion that ##na_n = ma_n## for all ##n\ge m## doesn't follow. Try again.
 
  • #3
1,456
44
That argument doesn't work. Given ##\epsilon > 0## there exists ##N## such that...
The problem is that the ##N## depends on ##\epsilon##. So the (obviously false) conclusion that ##na_n = ma_n## for all ##n\ge m## doesn't follow. Try again.
I guess that conclusion is obviously false when I look at it now... So I have that ##|(n-m)a_n|<\epsilon##. How can I get rid of that ##m## to get my result? Is there something simple that I am overlooking?
 

Related Threads on If a series converges with decreasing terms, then na_n -> 0

Replies
4
Views
1K
Replies
3
Views
374
Replies
18
Views
811
Replies
6
Views
3K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
17
Views
2K
Top