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Math Help - Find intervals of convergence

  1. #1
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    Find intervals of convergence

    I have a problem where the interval of convrsion is (0,2]
    However, for the end points plugging the x=0, and x=2 back in is where i am having prolems.
    I cannot figure it out what to do after i plug the x in. Can you someone help?

    f(x) =\sum_{n=0}^\infty  \frac {(-1)^{n+1}(x-1)^{n+1}} { n+1}

    I will attempt to solve this problem using the ratio


    f(x) =\sum_{n=0}^\infty  \frac {(-1)^{n+2}(x-1)^{n+2}} {n+2} . \frac { n+1} {(-1)^{n+1}(x-1)^{n+1}}

    I will attempt to properly cancel out.

    \lim_{n\to\infty} { |   (x-1)|}  < 1


    then i get for an answer 0,2

    so i plug x = 2 in to the original equation


    f(x) =\sum_{n=0}^\infty  \frac {(-1)^{n+1}(2-1)^{n+1}} { n+1}

    Which is
    f(x) =\sum_{n=0}^\infty  \frac {(-1)^{n+1}(1)^{n+1}} { n+1}

    From there i dont know what to do!
    Can anyhelp me finish this and do the x = 0 also?

    Thank you
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  2. #2
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    Re: Find intervals of convergence

    Quote Originally Posted by icelated View Post
    I have a problem where the interval of convrsion is (0,2]
    However, for the end points plugging the x=0, and x=2 back in is where i am having prolems.
    I cannot figure it out what to do after i plug the x in. Can you someone help?

    f(x) =\sum_{n=0}^\infty  \frac {(-1)^{n+1}(x-1)^{n+1}} { n+1}

    I will attempt to solve this problem using the ratio


    f(x) =\sum_{n=0}^\infty  \frac {(-1)^{n+2}(x-1)^{n+2}} {n+2} . \frac { n+1} {(-1)^{n+1}(x-1)^{n+1}}

    I will attempt to properly cancel out.

    \lim_{n\to\infty} { |   (x-1)|}  < 1


    then i get for an answer 0,2

    so i plug x = 2 in to the original equation


    f(x) =\sum_{n=0}^\infty  \frac {(-1)^{n+1}(2-1)^{n+1}} { n+1}

    Which is
    f(x) =\sum_{n=0}^\infty  \frac {(-1)^{n+1}(1)^{n+1}} { n+1}

    From there i dont know what to do!
    Can anyhelp me finish this and do the x = 0 also?

    Thank you
    To show the series is divergent when \displaystyle \begin{align*}x = 0  \end{align*}, substitute \displaystyle \begin{align*} x = 0 \end{align*} into your series

    \displaystyle \begin{align*} f(x) &= \sum_{n = 0}^{\infty}\frac{(-1)^{n+1}(x-1)^{n+1}}{n + 1} \\ f(0) &= \sum_{n = 0}^{\infty}\frac{(-1)^{n + 1}(0-1)^{n+1}}{n + 1} \\ &= \sum_{n = 0}^{\infty}\frac{(-1)^{n + 1}(-1)^{n + 1}}{n + 1} \\ &= \sum_{n = 0}^{\infty}\frac{\left[(-1)^{n + 1}\right]^2}{n + 1} \\ &= \sum_{n = 0}^{\infty}\frac{1}{n + 1} \end{align*}

    This is the Harmonic Series, which is well known to be (and very easy to show that it is) divergent.
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  3. #3
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    Re: Find intervals of convergence

    Thank you
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