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Math Help - Infinite Series Problem

  1. #1
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    Infinite Series Problem

    Having trouble with this problem, infact this whole topic, but thats another story

    The Nth partial sum, and the convergence behaviour, of the telescoping series


    [SIZE=+ 3]
    n = 1
    2 (2n + 1)(2n + 3)
    [/SIZE]
    how do i go about solving this? any help would be appreciated,

    thanks
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by joeseinfeld View Post
    Having trouble with this problem, infact this whole topic, but thats another story

    The Nth partial sum, and the convergence behaviour, of the telescoping series


    [SIZE=+ 3]
    n = 1
    2 (2n + 1)(2n + 3)
    [/SIZE]
    how do i go about solving this? any help would be appreciated,

    thanks
    I'm sorry to say, but I think its hard to decipher what the series in question is. That's why we like our members to learn some basic \text{\LaTeX} (see the attachment to this post)

    By any chance, could it possibly be \displaystyle\sum\limits_{n=1}^{\infty}\frac{1}{2(  2n+1)(2n+3)} or \displaystyle\sum\limits_{n=1}^{\infty}\frac{2}{(2  n+1)(2n+3)}?

    Please inform me which one you mean, and I'll fix it in your original post.
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    I'm sorry to say, but I think its hard to decipher what the series in question is. That's why we like our members to learn some basic \text{\LaTeX} (see the attachment to this post)

    By any chance, could it possibly be \displaystyle\sum\limits_{n=1}^{\infty}\frac{1}{2(  2n+1)(2n+3)} or \displaystyle\sum\limits_{n=1}^{\infty}\frac{2}{(2  n+1)(2n+3)}?

    Please inform me which one you mean, and I'll fix it in your original post.

    Yes it is the second one with the 2 ontop. sorry i didn't notice it came out like that.

    Thanks again
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  4. #4
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    Quote Originally Posted by joeseinfeld View Post
    Yes it is the second one with the 2 ontop.
    Hint: \displaystyle \sum_{n=1}^{\infty}\frac{2}{(2n+1)(2n+3)} = \sum_{n=1}^{\infty}\left(\frac{1}{2n+1}-\frac{1}{2n+3}\right) = \sum_{n=1}^{\infty}\frac{1}{2n+1}-\sum_{n=1}^{\infty}\frac{1}{2n+3}.
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by TheCoffeeMachine View Post
    Hint: \displaystyle \sum_{n=1}^{\infty}\frac{2}{(2n+1)(2n+3)} = \sum_{n=1}^{\infty}\left(\frac{1}{2n+1}-\frac{1}{2n+3}\right) = \sum_{n=1}^{\infty}\frac{1}{2n+1}-\sum_{n=1}^{\infty}\frac{1}{2n+3}.
    I would actually consider a finite series first and focus on the expansion of \displaystyle\sum_{n=1}^k\left(\frac{1}{2n+1}-\frac{1}{2n+3}\right) because its easier to see the numerous cancellations. Then, the OP should take k\to\infty in that result to complete the problem.
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  6. #6
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    Quote Originally Posted by Chris L T521 View Post
    I would actually consider a finite series first and focus on the expansion of \displaystyle\sum_{n=1}^k\left(\frac{1}{2n+1}-\frac{1}{2n+3}\right) because its easier to see the numerous cancellations.
    I don't really see how the cancellations are easier to see in the finite case.
    If anything, that seems unnecessary to me. But perhaps I'm missing something.
    Spoiler:
    \displaystyle\sum_{n=1}^{\infty}\frac{1}{2n+1} = \frac{1}{3}+\sum_{n=1}^{\infty}\frac{1}{2n+3}
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  7. #7
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    Quote Originally Posted by TheCoffeeMachine View Post
    I don't really see how the cancellations are easier to see in the finite case.
    If anything, that seems unnecessary to me. But perhaps I'm missing something.
    Spoiler:
    \displaystyle\sum_{n=1}^{\infty}\frac{1}{2n+1} = \frac{1}{3}+\sum_{n=1}^{\infty}\frac{1}{2n+3}
    These series are divergent to infinity.
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  8. #8
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    The series TheCoffeeMachine shows in his post are divergent but that is not the original problem. What do you get if you subtract \sum_{n=1}^\infty \frac{1}{2n+3} from that?

    Chris L T521's point with "consider finite series first" was that if n= 1, the sum \frac{1}{3}- \frac{1}{5}. If n= 2, \frac{1}{3}- \frac{1}{5}+ \frac{1}{5}- \frac{1}{7}= \frac{1}{3}- \frac{1}{7}, if n= 3, \frac{1}{3}- \frac{1}{5}+ \frac{1}{5}- \frac{1}{7}+ \frac{1}{7}- \frac{1}{9}= \frac{1}{3}- \frac{1}{9}, etc.
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  9. #9
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    Quote Originally Posted by Chris L T521 View Post
    I would actually consider a finite series first and focus on the expansion of \displaystyle\sum_{n=1}^k\left(\frac{1}{2n+1}-\frac{1}{2n+3}\right) because its easier to see the numerous cancellations.
    instead of doing that, having \dfrac1{2n+1}-\dfrac1{2n+3}, consider the first term, if we make n\to n+1 that yields the second term, hence, telescoping series and that's all.
    Last edited by Krizalid; October 12th 2010 at 02:40 PM.
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  10. #10
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    thanks guys figured it out after reading your replies

    i got Sn=(1/3)-1(2N+3).
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