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Math Help - Convergence of power series (d'Alemberts criterion)

  1. #1
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    Convergence of power series (d'Alemberts criterion)

    So, simply, where do I go wrong?


    The question:
    The interval of the power series \sum ^\infty _{n=1}\frac{(ax)^n}{n^2} is \left[ -\frac{1}{3},\frac{1}{3} \right] . Find a.

    My solution:
    For the series to be convergent, d'Alemberts ratio test must be satisfied, i.e. \lim_{n\rightarrow \infty} |\frac{u_{n+1}}{u_n}|<1 must hold true.

    For the series given we have:
    \lim_{n\rightarrow \infty} |\frac{u_{n+1}}{u_n}|=\lim_{n\rightarrow \infty} |\frac{(ax)^{n+1}n^2}{(n+1)^2(ax)^n}|
    = \lim_{n\rightarrow\infty}\frac{n^2}{(n+1)^2}|ax|=|  ax|=|a||x|<1\Leftrightarrow |x|<\frac{1}{|a|} .

    This gives -\frac{1}{|a|}<x<\frac{1}{|a|}

    It is easy to show that the series is convergent also for x=\frac{1}{|a|} and x=-\frac{1}{|a|} (by making these substitutions into the series given).

    So now we have that -\frac{1}{|a|}\leq x \leq \frac{1}{|a|} for the series to be convergent.

    But the interval of convergence is \left[ -\frac{1}{3},\frac{1}{3} \right] , meaning that a= \pm 3.


    However, the key in my book claims the answer to be a=\frac{1}{3}
    .



    Thank you in advance
    Aliquantus
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  2. #2
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    If it were the case that a = \frac {1}{3} then the interval of convergence is \left[ { - 3,3} \right].
    Thus something is amiss with the problem or the given answer.
    Review the statement and remember given answers are often incorrect.
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  3. #3
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    Thank you for your reply, Plato. I just got another reply on a swedish forum, also saying that there is a typo in my book. The only strange thing is that they give the same, apparently wrong, answer in both the key and the solution manual. I suppose not even authors are perfect. Thanks again!
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  4. #4
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    Quote Originally Posted by Aliquantus View Post
    Thank you for your reply, Plato. I just got another reply on a swedish forum, also saying that there is a typo in my book. The only strange thing is that they give the same, apparently wrong, answer in both the key and the solution manual. I suppose not even authors are perfect. Thanks again!

    You really shouldn't post the same questions on different forums. It considered impolite as someone could be spending their time replying to your question oblivious of the fact that it has already been answered elsewhere.

    Bobak
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