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Thread: Intervals of Convergence

  1. #1
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    Intervals of Convergence

    Find the intervals of convergence of f(x), f'(x) f''(x) and $\displaystyle \int {f(x)dx}$ for the function $\displaystyle f(x) = \sum_{n=1}^{\infty} ((-1)^{(n+1)} (x-2)^n) / n $ Check for convergence at endpoints of each inverval.

    I've been having a lot of trouble with problem and I'd really appreciate any help I can get.
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  2. #2
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    For f(x), apply the ratio test and find the limit as $\displaystyle n\rightarrow{\infty}$. Then check the endpoints as well. It converges at x=2.

    By the ratio test:

    $\displaystyle \frac{(-1)^{n+2}(x-2)^{n+1}}{n+1}\cdot\frac{n}{(-1)^{n+1}(x-2)^{n}}=\frac{n(2-x)}{n+1}$

    $\displaystyle \lim_{n\rightarrow{\infty}}\frac{n(2-x)}{n+1}=|2-x|$

    The ratio test gives cionvergence for $\displaystyle |2-x|<1$ and divergence for $\displaystyle |2-x|>1$

    Solving the first inequality we get $\displaystyle 1<x<3$.

    At x=1, the series is $\displaystyle -\sum_{n=1}^{\infty}\frac{1}{n}$, which is a divergent p-series.

    At x=3, it is convergent and converges to -ln(2).

    So, the convergence set is $\displaystyle 1<x\leq{3}$

    Check it out. Make sure I did not err somewhere, then run a test on the derivatives of the series.
    Last edited by galactus; May 3rd 2008 at 09:25 AM.
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  3. #3
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    Quote Originally Posted by galactus View Post
    At x=3, it is convergent and converges to -ln(2).
    What test did you use to find that?

    Check it out. Make sure I did not err somewhere, then run a test on the derivatives of the series.
    How would you approach the other three?
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  4. #4
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    Also, does anybody know how to integrate f(x)? We tried on a TI-89 but even that didn't give a definite answer.
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  5. #5
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    You can use the alternating series test to show convergence.

    But if x=3, you get $\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}$

    $\displaystyle 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+..............$

    This is a series I knew converges. This is the series associated with

    the "Euler's constant". Which shows $\displaystyle 1=ln(2)+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-.............$

    As for the antiderivative of your series. Integrating wrt x, wouldn't it be

    $\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n+1}(x-2)^{n+1}}{n(n+1)}$?.
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