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Math Help - [SOLVED] Convergence interval

  1. #1
    Bop
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    [SOLVED] Convergence interval

    Here is the serie:

    \sum{(-1)^{n-1}x^{n}(n/2n+1)^n}

    I don't get to find the convergence interval, I try with the alternating series test but I can't find it..

    Thank you!
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  2. #2
    MHF Contributor

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    Quote Originally Posted by Bop View Post
    Here is the serie:

    \sum{(-1)^{n-1}x^{n}(n/2n+1)^n}

    I don't get to find the convergence interval, I try with the alternating series test but I can't find it..

    Thank you!
    The alternating series test won't help fiind "interval of convergence" because this is not necessarily an "alternating series"! Specifically, for x negative, every term will be negative and not "alternating". (It is alternating for x positive, but you need to find all values of x for which it converges.)

    A power series always converges absolutely inside its radius of convergence and the standard method of finding that is to apply the "ratio test" or the "root test" to the absolute value of the series. Normally, the ratio test is easier to apply but those powers of n make this ideal for the root test!

    |a_n|= \frac{n^n}{(2n+1)^n} |x|^n so the nth root is |a_n|^{1/n}= \frac{n}{2n+1} |x|. This will converge absolutely if and only if the limit of that, as n goes to infinity, is less than 1:
    \lim_{n\to \infty}\frac{n}{2n+ 1}|x|= (1/2) |x|< 1.

    Now it should be clear what the radius of convergence and the interval of convergence is.
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