Does Cauchy's Theorem Confirm That Arithmetic Means of Null Sequences Are Null?

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Discussion Overview

The discussion revolves around the proof from Konrad Knopp's book regarding whether the arithmetic means of a null sequence also form a null sequence. Participants are examining the proof's details, particularly the role of the chosen natural number m in relation to ε and the implications for the sequence's convergence.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions how the chosen m in the proof can be considered fixed when it seems to depend on ε, suggesting that it changes with different ε values.
  • Another participant agrees that m is not fixed and varies, indicating that the numerator in the proof is also not a fixed number but is determined by m.
  • There is a discussion about the nature of ε in proofs, with a participant suggesting that ε can be treated as a fixed positive number during the proof process.
  • A participant seeks clarification on the last part of the proof after the inequality, indicating some uncertainty about the reasoning involved.

Areas of Agreement / Disagreement

Participants express differing views on the nature of m and its implications for the proof, indicating that there is no consensus on whether the numerator can be considered fixed or how it relates to ε. The discussion remains unresolved regarding the clarity of the proof's final steps.

Contextual Notes

Participants highlight potential limitations in understanding the proof, particularly regarding the dependence of m on ε and the implications for the convergence of the sequence.

elliti123
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Hello , i was just wondering if anyone could clarify one thing in this proof (its from Konrad Knopp book on infinite series) : If (x0,x1,...) is a null sequence, then the arithmetic means

xn'= x0+x1+x2+...+x/n+1 (n=1,2,3,...)

also forms a null sequence.

Proof: If ε >0 is given, then m can be so chosen, that for every n > m we have |xn| < ε/2 . For these n's, we have
|xn'| ≤ |x1+x2+x3+...+xm| / n+1 +(ε/2) (n-m /n+1)

since the numerator of the first fraction on the right hand side now contains a fixed number, we can further determine n0, so that for n > n0 that fraction remains < ε/2. But then, for every n > n0 , we have |xn'| < ε and our theorem is proved.

My question is: the chosen m in the proof as far as i know is a natural number changing according to what epsilon we give it so for example if the chosen m is 3 it might work for a particular ε but might not for another ε less than the other ε we have chosen first . So i have come to a conclusion that m or n0 that every n should be more than so the sequence converges to a real number is a function of epsilon therefore it changes whenever epsilon does. Now how exactly is the numerator they describe in the proof a fixed number?
Since the m changes whenever ε does then it is logical to infer that the summation of those terms would obviously change. And would you please explain the last part of the proof after the inequality i seem to have some vivid idea but i don't think i still get the last part. Thanks.
 
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##\epsilon## is given, you choose a different m and n each time.
 
Oh sorry, the one thing you wanted to ask is how the numerator on the left is a fixed number. It isn't but m determines it. M is different each time. So that numerator is different each time. But you choose it to have a certain property, that everything else in the sequence is small. Then you make n large to make that fraction small. Then everything is small.
 
verty said:
Oh sorry, the one thing you wanted to ask is how the numerator on the left is a fixed number. It isn't but m determines it. M is different each time. So that numerator is different each time. But you choose it to have a certain property, that everything else in the sequence is small. Then you make n large to make that fraction small. Then everything is small.
Haha there is a lot of small going on there.Anyways jokes aside so in this kind of proofs , i mean in general for epsilon proofs you actually do consider the epsilon you choose or give or even the epsilon itself to be a "fixed" positive number right?
 
What Limit is all about, the value we found after evaluating limit what this value actually shows.
 

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