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

In summary: 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...
  • #1
mertcan
345
6
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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  • #2
Hi, where is your attachment? ...
 
  • #3
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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  • #4
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.
 
  • #5
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?
 
  • #6
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??
 
  • #7
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.
 
  • #8
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 ?
 
  • #9
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?
 
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  • #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.
 

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

1. What is the power series ratio test?

The power series ratio test is a mathematical tool used to determine the convergence or divergence of an infinite series. It is specifically used for power series, which are infinite series of the form ∑(an(x-c)n), where an are coefficients and c is a constant.

2. How does the power series ratio test work?

The power series ratio test works by taking the limit of the ratio between consecutive terms in the series. If this limit, known as the ratio test, is less than 1, the series is convergent. If it is greater than 1, the series is divergent. If it is equal to 1, the test is inconclusive and another method must be used to determine convergence or divergence.

3. What is the formula for the power series ratio test?

The formula for the power series ratio test is as follows:

lim |an+1 / an| = L

If L < 1, the series is convergent. If L > 1, the series is divergent. If L = 1, the test is inconclusive.

4. Can the power series ratio test be used for any power series?

No, the power series ratio test can only be used for a specific type of power series known as a geometric series. This is a series where the ratio between consecutive terms is a constant, such as the series 1 + x + x2 + x3 + ... , where the ratio between terms is always x. For other types of power series, different convergence tests must be used.

5. What are the advantages of using the power series ratio test?

The power series ratio test is a simple and efficient way to determine the convergence or divergence of a geometric series. It is also a powerful tool for determining the radius of convergence of a power series, which is the range of values for which the series will converge. Additionally, the test is relatively easy to apply and can provide quick results for determining the convergence or divergence of a series.

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