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Math Help - Radius of convergence / endpoints

  1. #1
    VkL
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    Radius of convergence / endpoints

    I am really having trouble with this problem. The n'th term is given in the problem,


    (a) Determine the radius of convergence of the series.
    (b) What are the endpoints of the interval of convergence?
    left endpoint = x =
    right endpoint = x =
    Does the series converge at the left endpoint?
    Does the series converge at the right endpoint?



    I did the ratio test and got [(-1^n)*(x^n+1) / (n+1)] * [n / -1^(n-1) * x^n]

    I get infinity when i get an x in the numerator when i take the limit. Usually the x's cancel out. is this a special case? maybe i am doing something wrong, or don't know something. And I have no clue on how to find the end points even.

    Please help. Thank you.
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  2. #2
    MHF Contributor
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    Quote Originally Posted by VkL View Post
    I am really having trouble with this problem. The n'th term is given in the problem,


    (a) Determine the radius of convergence of the series.
    (b) What are the endpoints of the interval of convergence?
    left endpoint = x =
    right endpoint = x =
    Does the series converge at the left endpoint?
    Does the series converge at the right endpoint?



    I did the ratio test and got [(-1^n)*(x^n+1) / (n+1)] * [n / -1^(n-1) * x^n]

    I get infinity when i get an x in the numerator when i take the limit. Usually the x's cancel out. is this a special case? maybe i am doing something wrong, or don't know something. And I have no clue on how to find the end points even.

    Please help. Thank you.
    \lim_{n \to \infty} \left|\frac{(-1)^n x^{n+1}}{n+1} \cdot \frac{n}{(-1)^{n-1} x^n}\right| < 1<br />

    since we're working with the absolute value of the ratio, the (-1) factors can be ignored. simplifying ...

    \lim_{n \to \infty} \left|\frac{x \cdot n}{n+1}\right| < 1

    |x| \lim_{n \to \infty} \frac{n}{n+1} < 1

    |x| \cdot 1 < 1

    -1 < x < 1

    I leave you to check the endpoints for possible convergence.
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  3. #3
    VkL
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    on step 3, why was the absolute value of x taken out of the limit?
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  4. #4
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by VkL View Post
    on step 3, why was the absolute value of x taken out of the limit?
    Because n>0, so it is irrelevant.

    For the endpoints, plugging in 1 gives you \sum_{n=1}^{\infty}\frac{(-1)^n}{n}. This is the alternating harmonic series, so what can we say about it?

    Plugging in -1 gives you \sum_{n=1}^{\infty}\frac{-1}{n}. This is the negative harmonic series, so what can we say about it?

    Spoiler:
    The first one converges (to \ln 2) and the second one diverges. So this power series converges on x=(-1,1].
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