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Math Help - Telescoping Series

  1. #1
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    Telescoping Series

    Can somebody help with this?

    Evaluate as a telescoping series: The summation from k=1 to infinity of (k-1)/2^(k+1).

    The answer is 1/2 but I need to see the workings so I can figure out how it was arrived at.
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  2. #2
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    Quote Originally Posted by p75213 View Post
    Can somebody help with this?

    Evaluate as a telescoping series: The summation from k=1 to infinity of (k-1)/2^(k+1).

    The answer is 1/2 but I need to see the workings so I can figure out how it was arrived at.
    Hi

    \frac{k-1}{2^{k+1}} = \frac{k}{2^k}-\frac{k+1}{2^{k+1}}
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  3. #3
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    Quote Originally Posted by running-gag View Post
    Hi

    \frac{k-1}{2^{k+1}} = \frac{k}{2^k}-\frac{k+1}{2^{k+1}}
    Would I be right in saying you should also show that \lim_{k \to \infty} \frac{k-1}{2^k} = 0 (trivial really) since otherwise the telescoping thing might not work...
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  4. #4
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    Quote Originally Posted by Deadstar View Post
    Would I be right in saying you should also show that \lim_{k \to \infty} \frac{k-1}{2^k} = 0 (trivial really) since otherwise the telescoping thing might not work...
    I was just giving a hint to start with

    Since \frac{k-1}{2^{k+1}} = \frac{k}{2^k}-\frac{k+1}{2^{k+1}}

    \sum_{k=1}^{N} \frac{k-1}{2^{k+1}} = \sum_{k=1}^{N} \left(\frac{k}{2^k}-\frac{k+1}{2^{k+1}}\right) = \frac12 - \frac{N+1}{2^{N+1}}

    which converges towards \frac12 as N approaches infinity
    Last edited by running-gag; April 14th 2010 at 10:51 AM. Reason: English mistake
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  5. #5
    Super Member Deadstar's Avatar
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    Quote Originally Posted by running-gag View Post
    I was just given a hint to start with

    Since \frac{k-1}{2^{k+1}} = \frac{k}{2^k}-\frac{k+1}{2^{k+1}}

    \sum_{k=1}^{N} \frac{k-1}{2^{k+1}} = \sum_{k=1}^{N} \left(\frac{k}{2^k}-\frac{k+1}{2^{k+1}}\right) = \frac12 - \frac{N+1}{2^{N+1}}

    which converges towards \frac12 as N approaches infinity
    Lol yeah I know. I've been reading about the whole telescoping sum thing recently and it can easily be wrongly applied to infinite sums to show things like 0=1.

    Was just being annoyingly rigorous and hoping OP would catch on and realize either you must show it \to 0 or do it your way with only summing to N terms.
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  6. #6
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    You are right to be rigorous !
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  7. #7
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    Hi Guys,
    This is the part I need explained:-

    (k-1)/2^(k+1) = k/2^k-k+1/2^(k+1).
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  8. #8
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    \frac{k}{2^k} = \frac{2k}{2^{k+1}} (multiplying numerator and denominator by 2)

    \frac{k}{2^k}-\frac{k+1}{2^{k+1}} =  \frac{2k}{2^{k+1}}-\frac{k+1}{2^{k+1}} = \frac{2k-(k+1)}{2^{k+1}} = \frac{k-1}{2^{k+1}}
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  9. #9
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    How would we handle k-1/3^(k+1)? 2k/3^(K+1) won't break down into k/3^k.

    If we try the following: 3k/3^(k+1) - 2k+1/3^(k+1) = k/3^k - 2k+1/3^(k+1). Which is no longer a telescoping series ie. the middle bits don't cancel.
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  10. #10
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    \frac{k-1}{3^{k+1}} = \frac{2k-1}{4 \cdot 3^k} - \frac{2(k+1)-1}{4 \cdot 3^{k+1}}
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