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Math Help - Someone please help me with this seemingly geometric series

  1. #1
    s3a
    s3a is offline
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    Someone please help me with this seemingly geometric series

    Question:
    To show that the following
    http://www.wolframalpha.com/input/?i=sum+of+(k*(2%2F3)^k)+from+1+to+inf
    is convergent conditionally or absolutely.

    What I am having trouble with is the fact that it is multiplied by k. I realize that exponentiating something to k will grow faster than multiplying it by k but that is not sufficient at all for me to prove that this series converges. Actually I just used the ratio test now and found 2/3 which converges. So now a new question arises; is the ratio test valid for geometric series? Is this series a geometric one even though it is multiplied by k?

    Any input would be greatly appreciated!
    Thanks in advance!
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by s3a View Post
    Question:
    To show that the following
    http://www.wolframalpha.com/input/?i=sum+of+(k*(2%2F3)^k)+from+1+to+inf
    is convergent conditionally or absolutely.

    What I am having trouble with is the fact that it is multiplied by k. I realize that exponentiating something to k will grow faster than multiplying it by k but that is not sufficient at all for me to prove that this series converges. Actually I just used the ratio test now and found 2/3 which converges. So now a new question arises; is the ratio test valid for geometric series? Is this series a geometric one even though it is multiplied by k?

    Any input would be greatly appreciated!
    Thanks in advance!
    You may want to first derive a closed form for the series f(x) := \sum_{k=1}^\infty k x^k

    This can be done like this
    f(x)=x\cdot\sum_{k=1}^\infty k x^{k-1}=x\cdot\sum_{k=0}^\infty \frac{d}{dx} x^k

    =x\cdot \frac{d}{dx} \sum_{k=0}^\infty x^k=x\cdot \frac{d}{dx}\frac{1}{1-x}=\frac{x}{(1-x)^2}
    Now just plug in 2/3 for x, to get f(2/3)=\tfrac{2}{3}\cdot\frac{1}{\left(1-\frac{2}{3}\right)^2}=6.
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