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Thread: Powerseries - need help

  1. #1
    Sep 2007

    Powerseries - need help

    Find the convergate-area to the following series:

    SUM (-1)^n * x^n / n(n-1)

    SUM n / 2^n * (x-1)^n

    SUM (-1)^n * 4n * x^n / n ln n

    Struggling a bit with these, so I would highly appreciate ANY help, thanks.
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  2. #2
    Junior Member
    Aug 2007

    SUM (-1)^n * x^n / n(n-1)

    Im just learning this topic so bear with me.

    You can solve this by using the ratio test.

    Absolute value of A(n+1)/A(n)
    The Absoulte value of (-1)^n goes away since its just 1.

    so its:

    x^n+1 -----n(n-1)
    ------- * ---------
    n+1(n) -----x^n

    cross cancel you get

    x limit of n -> infinity (n-1)/(n+1) which ='s 1

    abs of x < 1 = -1 < x < 1

    then plug -1 and 1 in and test if they converge of not.

    The 2nd problem you can solve by doing the root test.

    SUM n / 2^n * (x-1)^n

    if you root the equation to the n, the power n cancel out

    = n/2 * x-1

    you can move x-1 outside of the series and have lim n -> infinity n/2 which = infinity only if x is not 1.

    If anyone who can double check me would be great. Again, Im not 100% sure, Im still learning this topic.

    The third one you can do by using the ratio test again. Try it.
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  3. #3
    Senior Member
    Dec 2007
    Anchorage, AK

    Re: Power Series

    Actually, ff4930, it looks like you made an error in applyring the root test to the second series:
    $\displaystyle \sqrt[n]{|a_n|}=\sqrt[n]{|\dfrac{n}{2^n}(x-1)^n|}$
    $\displaystyle =\dfrac{\sqrt[n]{n}}{2}|x-1|$

    not the $\displaystyle \dfrac{n}{2}|x-1|$ that you have, and
    $\displaystyle \lim_{n \to \infty}\sqrt[n]{n} = 1$, so that gives |x-1|<2 for the area of convergence. In fact, this problem is done more easily with the ratio test:
    $\displaystyle \left | \frac{a_{n+1}}{a_n} \right | = \left | \cfrac{\cfrac{(n+1)(x-1)^{n+1}}{2^{n+1}}}{\cfrac{n(x-1)^n}{2^n}} \right |$
    $\displaystyle = \dfrac{(n+1)|x-1|}{2n}$

    and $\displaystyle \lim_{n \to \infty}\dfrac{n+1}{2n} = \tfrac{1}{2}$.

    Warrick2236, as for the third one, unless you typed it wrong, it looks as if you have an n in the numerator that can cancel with the n in the denominator.

    --Kevin C.
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