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Math Help - Calculating series sum

  1. #1
    Senior Member Twig's Avatar
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    Calculating series sum

    Hi!

    This one is giving me problems.

    \sum_{k=1}^{\infty}\, \frac{1}{(2k-1)(2k+1)} = \frac{1}{1\cdot 3} + \frac{1}{3\cdot 5}\, ...

    The book tells me to calculate the n:th partialsum by dividing a_{k} into two or more terms. Then to prove that the series converge and calculate its sum.

    What is it that I am supposed to do really, donīt really get it.

    Thanks
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  2. #2
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    Quote Originally Posted by Twig View Post
    Hi!

    This one is giving me problems.

    \sum_{k=1}^{\infty}\, \frac{1}{(2k-1)(2k+1)} = \frac{1}{1\cdot 3} + \frac{1}{3\cdot 5}\, ...

    The book tells me to calculate the n:th partialsum by dividing a_{k} into two or more terms. Then to prove that the series converge and calculate its sum.

    What is it that I am supposed to do really, donīt really get it.

    Thanks
    Partial fraction decomposition of a_k:


    a_k = \frac{1}{(2k-1)(2k+1)} = \frac{1}{2} \left( \frac{1}{2k-1} - \frac{1}{2k+1}\right).


    You should see what happens next ....
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  3. #3
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    \sum\limits_{k=0}^{\infty }{\frac{1}{(2k+1)(2k+3)}}=\frac{1}{3}+\sum\limits_  {k=1}^{\infty }{\frac{1}{(2k+1)(2k+3)}}, and from there the convergence is easy, since for each k\ge1 we have,

    \frac{1}{(2k+1)(2k+3)}=\frac{1}{4k^{2}+8k+3}\le \frac{1}{4k^{2}}, then your series converges by comparison with \sum\limits_{k=1}^{\infty }{\frac{1}{k^{2}}} which is a p-series for p=2>1.
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  4. #4
    Senior Member Twig's Avatar
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    hi

    Hi

    Thanks for your help guys, it became a bit clearer for me.
    I guess the 'trick' here is to write out some terms when a_{k}
    has been partial divided, and see what terms cancel out each other?

    I got:

    \frac{1}{2}(\frac{1}{2k - 1} - \frac{1}{2k + 1}) + \frac{1}{2}(\frac{1}{2k + 1} - \frac{1}{2k+3})\, ...

    I can see that some terms cancel each other.
    So I guess I`ll end up with:

    \frac{1}{2}(1 - \frac{1}{2n + 1}) \rightarrow \frac{1}{2}\; \mbox{when}\; n\rightarrow \infty

    Also thanks Krizalid for helping me with the convergence proof.
    One question however, why did you write

     \frac{1}{3} + \sum_{k=1}^{\infty}\, \frac{1}{(2k+1)(2k+3)} ?

    Couldnīt we just make proof from scratch, without explicitly writing out the first term \frac{1}{3} ?
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  5. #5
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    Quote Originally Posted by Twig View Post

    One question however, why did you write

     \frac{1}{3} + \sum_{k=1}^{\infty}\, \frac{1}{(2k+1)(2k+3)} ?
    'cause \sum\limits_{k=1}^{\infty }{\frac{1}{(2k-1)(2k+1)}}=\sum\limits_{k=0}^{\infty }{\frac{1}{(2k+2-1)(2k+2+1)}}=\sum\limits_{k=0}^{\infty }{\frac{1}{(2k+1)(2k+3)}}.

    Now I just considered the last sum and summed the first term to start the sum in k=1.
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  6. #6
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    Quote Originally Posted by Twig View Post
    Hi

    Thanks for your help guys, it became a bit clearer for me.
    I guess the 'trick' here is to write out some terms when a_{k}
    has been partial divided, and see what terms cancel out each other?

    I got:

    \frac{1}{2}(\frac{1}{2k - 1} - \frac{1}{2k + 1}) + \frac{1}{2}(\frac{1}{2k + 1} - \frac{1}{2k+3})\, ...

    I can see that some terms cancel each other.
    So I guess I`ll end up with:

    \frac{1}{2}(1 - \frac{1}{2n + 1}) \rightarrow \frac{1}{2}\; \mbox{when}\; n\rightarrow \infty

    Also thanks Krizalid for helping me with the convergence proof.
    One question however, why did you write

     \frac{1}{3} + \sum_{k=1}^{\infty}\, \frac{1}{(2k+1)(2k+3)} ?

    Couldnīt we just make proof from scratch, without explicitly writing out the first term \frac{1}{3} ?
    As an alternative:

    \frac{1}{(2k - 1)(2k + 1)} = \frac{1}{4k^2 - 1} \leq \frac{1}{4k^2 - k^2} = \frac{1}{3k^2} = \frac{1}{3} \cdot \frac{1}{k^2}

    and convergence is assured by the comparison test.
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