Another convergence. An interval of convergence.

I have to find, for all value of constant , the interval of converence of .

How do I do? By using the Ratio test I get

Now I will use the rule of Hopital to make this fraction 'simpler'.

According to the Ratio test, it states that this series is convergent if , that is . Divergent if , that is . If , it has not result. If we test it by inserting the value of on this series, it would be divergent according to Wolfram Alpha. I don't understand this part because this problem clearly states the value of a should be greater than 0, so should also be convergent.

The other hand if we see it converges if where , so I am not really sure I could say it would be . If it's true, the interval of convergent would be like as .

If I did it wrong, please tell me what to do.

Re: Another convergence. An interval of convergence.

Sorry, I won't analyze your solution and will just write what I think. According to this page, the radius of convergence is the reciprocal of . If , the limit is 1, so the radius is 1. If a > 1, the limit is 1 / a, so the radius is a.

Now, what happens when |x| equals the radius? If a < 1 and x = 1, the series diverges by the integral test. If a = x = 1, the series clearly diverges. If a > 1 and x = a, again the series diverges by the integral test. It is left to check when x equals negative radius.

Re: Another convergence. An interval of convergence.

Thanks. Didn't think about it before. I have tried to check it and you were right about the values of a and r.

Generally |x| should be lesser than r, so that the series converges. If greater than r, it diverges.

The other hand |x|/a < 1 the series converges which means |x|<a or -a < x < a. So what do I have to answer more presicely?

Re: Another convergence. An interval of convergence.

You have to realise that there are infinitely many possible series that as written, your expression represents (as r and a are freely chosen). Luckily the ratio test can test all but TWO of these cases, where the ratio's limit is 1. So these two series where the ratio's limit is 1 needs to be tested with some other test as Emakarov has done :)

Re: Another convergence. An interval of convergence.

Quote:

Originally Posted by

**MathsforNewbs** The other hand |x|/a < 1 the series converges which means |x|<a or -a < x < a. So what do I have to answer more presicely?

As I said in my previous post, the radius of convergence is not always . This is because is not always |x| / a. And this is because one of the conditions of the l'Hôpital's rule is that the numerator and the denominator must both tend to 0 or . But when a < 1, and tend to 1, so the l'Hôpital's rule does not apply. When a < 1, the limit above is |x|, so the radius of convergence is 1. Altogether, the radius is when a > 1 and 1 when a ≤ 1.

Also, it is easier to see that the interval of convergence is open in all cases not by the integral test, as I said in the previous post, but because when x equals one of the interval's edges, the terms of the series don't tend to zero.