Solving Convergent Series Problem: Proving Limit of nan is 0

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Discussion Overview

The discussion revolves around proving that the limit of the sequence \( n a_n \) approaches 0, given that \( a_n \) is a decreasing sequence of non-negative terms and that the series \( S_n \) converges. The focus is on exploring different approaches to establish this limit, including the use of bounding techniques and properties of convergent series.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant proposes starting with the sequence \( b_n = a_{n+1} + a_{n+2} + ... + a_{2n} \) and shows that the limit of \( b_n \) is 0.
  • Another participant suggests that the decreasing nature of the sequence implies the result for even \( n \), which could extend to all \( n \).
  • A participant expresses confusion about bounding \( n a_n \) from above, noting that they cannot use \( a_n \) itself since \( n a_n \geq a_n \).
  • It is mentioned that proving the limit for even \( n \) also leads to the conclusion for odd \( n \) due to the properties of decreasing sequences.

Areas of Agreement / Disagreement

Participants generally agree on the implications of the decreasing nature of the sequence but have not reached a consensus on the bounding techniques or the next steps in the proof.

Contextual Notes

There are limitations in the discussion regarding the assumptions made about bounding sequences and the specific mathematical steps that remain unresolved. The use of comparison tests and the squeeze theorem is mentioned but not fully explored.

steven187
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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 :confused:
 
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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.]
 
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 can't 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 can't 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 I am sure it would be pretty useful and quik
 
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.
 

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