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Math Help - telescopic series

  1. #1
    Senior Member sfspitfire23's Avatar
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    telescopic series

    Hi guys, quick Question

    For the series  \sum_{i=1}^\infty(\frac{1}{k}-\frac{1}{k+2}) find a formula for S_n

    I know that every number will cancel out but 1 and \frac{1}{2} but the answer is 1+\frac{1}{2}-\frac{1}{n+1}-\frac{1}{n+2}.

    How do we get -\frac{1}{n+1}-\frac{1}{n+2}?

    Thx!
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  2. #2
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    S_n is not the sum of the infinite series, but the partial sum \sum_{k=1}^n(\frac{1}{k}-\frac{1}{k+2}).

    You see it in the definition of the sum of an infinite series \sum_{k=1}^{\infty}(\frac{1}{k}-\frac{1}{k+2}) is defined as \lim_{n \to \infty}S_n, where S_n is as above.

    - Hollywood
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  3. #3
    Senior Member sfspitfire23's Avatar
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    Indeed. How do you get that partal sum? I understand the 1 and .5 part...
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  4. #4
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    Everything cancels out except for 1 and 1/2 at the beginning and 1/(k+1) and 1/(k+2) at the end:

    S_n=\sum_{k=1}^n(\frac{1}{k}-\frac{1}{k+2})=\sum_{k=1}^n\frac{1}{k}-\sum_{k=1}^n\frac{1}{k+2}=\sum_{k=1}^n\frac{1}{k}-\sum_{k=3}^{n+2}\frac{1}{k}

    S_n=1+\frac{1}{2}+\sum_{k=3}^n\frac{1}{k}-\sum_{k=3}^n\frac{1}{k}-\frac{1}{n+1}-\frac{1}{n+2}

    S_n=1+\frac{1}{2}-\frac{1}{n+1}-\frac{1}{n+2}

    - Hollywood
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  5. #5
    Senior Member sfspitfire23's Avatar
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    I guess I'm still not sure how to get the right side. How do I produce it in the first place? I understand the 1 and .5. Thanks for your patience.
    Last edited by sfspitfire23; March 22nd 2010 at 12:24 PM.
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  6. #6
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    \frac{1}{k}-\frac{1}{k+2}=\frac{1}{k}-\frac{1}{k+1}+\frac{1}{k+1}-\frac{1}{k+2}, there's two telescoping sums there.
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  7. #7
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    Quote Originally Posted by sfspitfire23 View Post
    I guess I'm still not sure how to get the right side. How do I produce it in the first place? I understand the 1 and .5. Thanks for your patience.
    If you look at the series this way:

    (1-\frac{1}{3})+(\frac{1}{2}-\frac{1}{4})+(\frac{1}{3}-\frac{1}{5})+(\frac{1}{4}-\frac{1}{6})+...

    the partial sums S_n are:

    S_n=(1-\frac{1}{3})+(\frac{1}{2}-\frac{1}{4})+(\frac{1}{3}-\frac{1}{5})+(\frac{1}{4}-\frac{1}{6})+...+(\frac{1}{n-1}-\frac{1}{n+1})+(\frac{1}{n}-\frac{1}{n+2})

    It's no longer infinite, just the first n terms.

    So you can see that the 1/3 in the first term cancels the one in the third term, the 1/4 in the second term cancels the one in the fourth term, and so on. But when we get to 1/(n+1), there is no term to cancel it, and the same for 1/(n+2).

    Hope this helps - feel free to post again in this thread if you're still having trouble.

    - Hollywood
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