# I Power series ratio test

1. Jun 11, 2016

### mertcan

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....

2. Jun 12, 2016

### Ssnow

Hi, where is your attachment? ...

3. Jun 12, 2016

### mertcan

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. Jun 12, 2016

### Ssnow

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. Jun 12, 2016

### mertcan

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. Jun 12, 2016

### mertcan

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. Jun 12, 2016

### FactChecker

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. Jun 12, 2016

### mertcan

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. Jun 12, 2016

### FactChecker

I mean that only for the case of x = +-L. It works otherwise.

10. Jun 12, 2016

### mertcan

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: Jun 12, 2016
11. Jun 12, 2016

### FactChecker

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.