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Homework Help: Interesting Telescoping Series - Calc 2

  1. Sep 16, 2011 #1
    Interesting Telescoping Series -- Calc 2

    My problem with this series arose when I attempted a partial fractions decompisition of the following, (k-1)/(2^(2k+1). I attempted to factor the denominator with 2(2^k) which is right, but where do I go from there? It does not help much to multiply A*(2^k).

    My decomposition was {a*2^k+2b}/2(2^k), where did I go wrong?

    Thanks,

    Dane
     
  2. jcsd
  3. Sep 16, 2011 #2

    SammyS

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    Re: Interesting Telescoping Series -- Calc 2

    Why try partial fractions? I doubt that it is even applicable for this.

    Does your sum start at zero or does it start at 1 ?
     
  4. Sep 16, 2011 #3
    Re: Interesting Telescoping Series -- Calc 2

    Problem attached............


    It is a telescoping series from shaums outline

    Thanks alot
     

    Attached Files:

    Last edited: Sep 16, 2011
  5. Sep 16, 2011 #4

    SammyS

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    Re: Interesting Telescoping Series -- Calc 2

    So, the problem is to evaluate the series:
    [itex]\displaystyle \sum_{k=1}^{\infty}\frac{k-1}{2^{k+1}}[/itex]​
    (There's a typo in your Original Post or in your attached file.)

    Note that [itex]\displaystyle \sum_{k=1}^{\infty}\frac{k-1}{2^{k+1}}\ =\ \sum_{k=1}^{\infty}\frac{k}{2^{k+1}}\ -\ \sum_{k=1}^{\infty}\frac{1}{2^{k+1}}[/itex] Added in Edit: I doubt that this helps.
     
    Last edited: Sep 16, 2011
  6. Sep 16, 2011 #5

    SammyS

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    Re: Interesting Telescoping Series -- Calc 2

    The k = 1 term is zero.

    [itex]\displaystyle \sum_{k=1}^{\infty}\frac{k-1}{2^{k+1}}=0+\sum_{k=2}^{\infty}\frac{k-1}{2^{k+1}}[/itex]

    Let k = j+1. Rewrite the sum on the right in terms of j.

    [itex]\displaystyle \sum_{k=2}^{\infty}\frac{k-1}{2^{k+1}}=\sum_{j=1}^{\infty}\frac{j}{2^{j+2}}[/itex]
    [itex]\displaystyle =\frac{1}{2}\sum_{j=1}^{\infty}\frac{j-1+1}{2^{j+1}}[/itex]

    [itex]\displaystyle =\frac{1}{2}\left(\sum_{j=1}^{\infty}\frac{j-1}{2^{j+1}}+\sum_{j=1}^{\infty}\frac{1}{2^{j+1}} \right)[/itex]
    The first sum in the last line is the same as the beginning series. -- It's just written with j rather than k.

    Do some algebra to find that [itex]\displaystyle \sum_{k=1}^{\infty}\frac{k-1}{2^{k+1}}=0+\sum_{j=1}^{\infty}\frac{1}{2^{j+1}}[/itex]

    The sum with j is a geometric series. It converges to a well known result, but can also be evaluated as a telescoping series in a manner similar to what was done above.
     
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