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Math Help - Another convergence. An interval of convergence.

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    Another convergence. An interval of convergence.

    I have to find, for all value of constant a>0, the interval of converence of $\sum_{n=0}^{\infty}\frac{x^{n}}{1+a^{n}}$.
    How do I do? By using the Ratio test I get
    \lim_{n \to \infty}\frac{\left (\frac{x^{n+1}}{1+a^{n+1}}  \right )}{\left (\frac{x^{n}}{1+a^{n}}  \right )}=\lim_{n \to \infty} \frac{x^{n+1}}{x^{n}}\frac{1+a^{n}}{1+a^{n+1}}= \lim_{n \to \infty}x\frac{1+a^{n}}{1+a^{n+1}}
    Now I will use the rule of Hopital to make this fraction 'simpler'.
    \lim_{n \to \infty}x\frac{1+a^{n}}{1+a^{n+1}}=x\lim_{n \to \infty}\frac{\frac{\mathrm{d} \left (1+a^{n}  \right )}{\mathrm{d} n}}{\frac{\mathrm{d} \left (1+a^{n+1}  \right )}{\mathrm{d} n}}=...=x\lim_{n \to \infty}a^{-1}=\frac{x}{a}
    According to the Ratio test, it states that this series is convergent if \frac{x}{a}<1, that is x<a. Divergent if \frac{x}{a}>1, that is x>a. If x=a, it has not result. If we test it by inserting the value of a = 1 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 a = 1 > 0 should also be convergent.

    The other hand if we see it converges if x<a where a>0, so I am not really sure I could say it would be 0<x<a. If it's true, the interval of convergent would be like x \in ]0, a] as a > 0.
    If I did it wrong, please tell me what to do.
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    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 \lim_{n\to\infty}\left| \frac{1+a^n}{1+a^{n+1}} \right|. If a\le1, 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.
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    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?
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    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
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    Re: Another convergence. An interval of convergence.

    Quote Originally Posted by MathsforNewbs View Post
    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 a. This is because \lim_{n\to\infty} \left| x\frac{1+a^n}{1+a^{n+1}} \right| is not always |x| / a. And this is because one of the conditions of the l'H˘pital's rule is that the numerator and the denominator must both tend to 0 or \pm\infty. But when a < 1, 1+a^n and 1+a^{n+1} tend to 1, so the l'H˘pital's rule does not apply. When a < 1, the limit above is |x|, so the radius of convergence is 1. Altogether, the radius is a 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.
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