Is there a proof for the precision of convergence or divergence at x=+,-(1/L)?

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Discussion Overview

The discussion revolves around the convergence or divergence of a series at the critical points \( x = \pm \frac{1}{L} \). Participants explore whether there is a definitive proof regarding the precision of convergence or divergence at these points, examining the implications of the ratio test and the behavior of terms in the series.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants note that the critical case occurs when \( x = \pm \frac{1}{L} \), where the limit of the ratio of terms approaches 1, making it difficult to determine convergence.
  • Others express a desire for a mathematical proof that clarifies why convergence or divergence cannot be decided at these critical points.
  • Some participants argue that the ratio test is too general and can yield the same limit \( L \) for different series, leading to both converging and diverging examples.
  • A participant mentions that while there are proofs for convergence when \( |x| < L \) and divergence when \( |x| > L \), only counterexamples exist for \( |x| = L \), suggesting a lack of definitive proof for this case.
  • Questions arise about the implications of summing similar terms as \( n \) approaches infinity, with some participants seeking clarification on this reasoning.

Areas of Agreement / Disagreement

Participants generally agree that the case of \( x = \pm \frac{1}{L} \) requires special consideration, but there is no consensus on whether a definitive proof exists regarding convergence or divergence at these points. Multiple competing views remain regarding the applicability of the ratio test and the nature of convergence in this context.

Contextual Notes

Limitations include the dependence on the definitions of convergence and divergence, the generality of the ratio test, and the unresolved nature of mathematical proofs specific to the critical points.

mertcan
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hi, If you look at my attachment you can see that the book express that for the situation of x=+,-(1/L) we need further investigation. It means being converged or diverged is not precise. I would like to ask: Is there remarkable proof that if x=+,-(1/L) convergence or divergence is not precise? Could you provide me with that proof? I really really wonder it...Thanks in advance...
 
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Hi, where is your attachment? ...
 
ohh I really apologise... I upload it here, and looking forward to your response
Also let me express my question again: If you look at my attachment you can see that the book express that for the situation of x=+,-(1/L) we need further investigation. It means being converged or diverged is not precise. I would like to ask: Is there remarkable proof that if x=+,-(1/L) convergence or divergence is not precise? Could you provide me with that proof? I really really wonder it...Thanks in advance...
 

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The idea is that the critical case is when ##x=\pm \frac{1}{L}##, because in this case the comparison between terms in the sum as ##a_{n+1}x^{n+1}## and ##a_{n}x^{n}## have the same behavior for ##n\rightarrow \infty## (that is the limit of the ratio is ##1##), so if the behavior of every term is the same of the previous term at infinity you cannot decide if it converge or not... The case when ##x=\pm \frac{1}{L}## must be treat separately in order to decide if in this point the series converge or not. I hope to have clarify something.
 
Ssnow said:
The idea is that the critical case is when ##x=\pm \frac{1}{L}##, because in this case the comparison between terms in the sum as ##a_{n+1}x^{n+1}## and ##a_{n}x^{n}## have the same behavior for ##n\rightarrow \infty## (that is the limit of the ratio is ##1##), so if the behavior of every term is the same of the previous term at infinity you cannot decide if it converge or not... The case when ##x=\pm \frac{1}{L}## must be treat separately in order to decide if in this point the series converge or not. I hope to have clarify something.
Actually, I am aware that comparison between terms in the sum as ##a_{n+1}x^{n+1}## and ##a_{n}x^{n}## have the same behavior for ##n\rightarrow \infty## but my main question is : Why can't we decide whether it converges or not? Is there mathematical proof?
 
Also my second question is: I consider that when n is at infinity or very very large number, we have lots of same number and if we sum same numbers up we have infinity. Am I right??
 
We can't use that rule to decide if it converges because the rule is too general for that. Because L is defined using the absolute values of the ai, there are too many different cases that will give the same L. Some converge and others do not. You can make examples that do anything you want just by manipulating the signs of the coefficients, ai.
 
FactChecker said:
We can't use that rule to decide if it converges because the rule is too general for that. Because L is defined using the absolute values of the ai, there are too many different cases that will give the same L. Some converge and others do not. You can make examples that do anything you want just by manipulating the signs of the coefficients, ai.
You say that ratio test can not tell us whether or not it converges? I am confusing Could you spell it out giving some Mathematical Stuff, proofs derivations ?
 
mertcan said:
You say that ratio test can not tell us whether or not it converges? I am confusing Could you spell it out giving some Mathematical Stuff, proofs derivations ?
I mean that only for the case of x = +-L. It works otherwise.
 
  • #10
FactChecker said:
I mean that only for the case of x = +-L. It works otherwise.
Ok I got it if ratio is 1 the your example is satisfying, but I am also curious about the Mathematical proof of it , I need some proof to convince myself. Could you give some proof that at critics point we do not know convergence or divergence besides the examples?
 
Last edited:
  • #11
mertcan said:
Ok I got it if ratio is 1 the your example is satisfying, but I am also curious about the Mathematical proof of it , I need some proof to convince myself. Could you give some proof that at critics point we do not know convergence or divergence besides the examples?
There is a proof that it will converge if |x| < L and a proof that it will not converge if |x| > L. There are only counterexamples for the case |x| = L. The counterexamples prove that the proof does not work in that case.
 

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