How Does the Ratio Test Prove Series Convergence?

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Discussion Overview

The discussion revolves around the proof of the ratio test for series convergence, specifically focusing on the implications of the limit definition and the conditions under which the test concludes convergence. Participants explore the nuances of the proof and the assumptions involved.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion regarding the proof of the ratio test, particularly about the necessity of the condition that \(\frac{a_{n+1}}{a_{n}} < R\) for all \(n > N\) and its implications for the conclusion about \(a_{N+1} < a_{n}R\).
  • Another participant suggests that understanding the meaning of the limit's existence might clarify the proof, indicating that for sufficiently large \(n\), the ratio will be less than or equal to \(R\).
  • A participant questions the correctness of a specific part of the proof, noting that the limit definition typically uses a greater-than condition, which they find confusing.
  • One participant argues that the definitions of limits can be equivalent in terms of the conditions used, suggesting that both greater-than and greater-than-or-equal-to can be applied consistently.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of the limit definition and its application in the proof. There is no consensus on the correctness of the bold statement in the proof, and the discussion remains unresolved regarding the implications of the limit's conditions.

Contextual Notes

Participants highlight potential ambiguities in the proof related to the definitions of limits and the assumptions made about the sequence. The discussion does not resolve these ambiguities.

Bipolarity
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I am trying to understand something in the proof of the ratio test for series convergence.
If a_{n} is a sequence of positive numbers, and that the ratio test shows that \lim_{n→∞}\frac{a_{n+1}}{a_{n}} = r &lt; 1, then the series converges.

Apparently, the proof defines a number R : r<R<1, and then shows that there exists a N>0 such that \frac{a_{n+1}}{a_{n}} &lt; R for all n>N. It need not to be true in the case where n=N, right? Up to this part I get.

But then it concludes from the above that, there exists a positive N such that
a_{N+1}&lt;a_{n}R which does not follow due to the statement in bold.

Could someone please point out where I am wrong so I can continue this theorem without any qualms? Thanks!

BiP
 
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Hmm, think about what it means for the limit to exist. I think that might be the missing piece that shows why it works. Because if you believe it exist, there is an n greater than or equal to N such that (a_(n+1))/(a_n) is less than or equal to R for n greater than N. Then from there you can rewrite the inequality to get what you got.
 
MarneMath said:
Hmm, think about what it means for the limit to exist. I think that might be the missing piece that shows why it works. Because if you believe it exist, there is an n greater than or equal to N such that (a_(n+1))/(a_n) is less than or equal to R for n greater than N. Then from there you can rewrite the inequality to get what you got.

Are you sure the part in bold is correct? Doesn't the limit definition exclusively use greater than? Because that is precisely what I don't fully understand.

BiP
 
Bipolarity said:
Are you sure the part in bold is correct? Doesn't the limit definition exclusively use greater than? Because that is precisely what I don't fully understand.

BiP

It doesn't really matter. The statement:

For all \varepsilon&gt;0, there exists an N such that for all n\geq N holds that |a_n-a|&lt;\varepsilon.

is actually equivalent with

For all \varepsilon&gt;0, there exists an N such that for all n&gt; N holds that |a_n-a|&lt;\varepsilon.

So you can use both statements to define limit of a sequence. Of course, once you decided on which of both versions to use, you have to be consistent.
 

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