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Math Help - Sequence Help

  1. #1
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    Cool Sequence Help

    I'm studying for my final and my teacher gave me a worksheet. I have two sequence problems I do not understand at all. If anyone could please help I'd appreciate it!

    1. Find the sum of the sequence, note i=symbol for factorial, I couldn't find a factorial symbol :

    ((2n-1)i) / ((2n+1)i)


    2. find sum of sequence (That's exactly how written, I have no idea...) :

    2, -1/2 + 1/8 - ... 1/2048
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  2. #2
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    Try [shift]+1

    Have you noticed that \frac{(2n-1)!}{(2n+1)!} = \frac{1}{2n(2n+1)}?
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  3. #3
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    Quote Originally Posted by TKHunny View Post
    Try [shift]+1

    Have you noticed that \frac{(2n-1)!}{(2n+1)!} = \frac{1}{2n(2n+1)}?
    Never noticed that, how did you get it? and how does that lead to the sum?
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  4. #4
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    Expand the factorials. Only two terms differ.
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  5. #5
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    Quote Originally Posted by TKHunny View Post
    Expand the factorials. Only two terms differ.
    So when it is fully evaluated it would be 1 / (4n^2 + 2n)
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  6. #6
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    "fully evaluated" means nothing.

    Finding a parsimonious expression for a sum can be an art. Wonderful accolades have been awarded because of such discoveries. Other than self-awesomeness, though, you need a bag of references. In this case, notice further that:

    \frac{1}{2n(2n+1)}\;=\;\frac{1}{2n} - \frac{1}{2n+1}

    Write out the expression for n = 1, 2, and 3 and see if you notice anything. (Resist the temptation to add anything.)
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  7. #7
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    Quote Originally Posted by TKHunny View Post
    "fully evaluated" means nothing.

    Finding a parsimonious expression for a sum can be an art. Wonderful accolades have been awarded because of such discoveries. Other than self-awesomeness, though, you need a bag of references. In this case, notice further that:

    \frac{1}{2n(2n+1)}\;=\;\frac{1}{2n} - \frac{1}{2n+1}

    Write out the expression for n = 1, 2, and 3 and see if you notice anything. (Resist the temptation to add anything.)
    I don't quite understand what you're getting at. I made a mistake in the original post, it doesn't want the sum of the expression, but rather it says to evaluate it.

    also, anything on the second question?
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  8. #8
    Super Member TheChaz's Avatar
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    2, -1/2 + 1/8 - ... +1/2048
    This looks like a geometric sequence with first term "2" and ratio "-1/4"...
    Last edited by TheChaz; June 6th 2011 at 08:07 PM. Reason: sign!
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  9. #9
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    You probably should differentiate between a sequence (a list) and a series (add things up).

    Did you list the terms associated with n = 1, 2, and 3? Really, it will be quite enlightening. You will fail this course if you can't list three terms of a series.

    n = 1 leads to \frac{1}{2(1)} - \frac{1}{2(1)+1}\;=\;\frac{1}{2}-\frac{1}{3}

    How about n = 2?

    Note: That ratio on #2 looks like "-1/4" and it appears to be finite.
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  10. #10
    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by TacticalPro View Post
    Never noticed that, how did you get it? and how does that lead to the sum?
    \frac{1}{2n(2n+1)}=\frac{1}{2n}- \frac{1}{2n+1}
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  11. #11
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    Quote Originally Posted by TKHunny View Post

    Have you noticed that \frac{(2n-1)!}{(2n+1)!} = \frac{1}{2n(2n+1)}?
    Never noticed that, how did you get it? and how does that lead to the sum?
    Telescoping series - Wikipedia, the free encyclopedia
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  12. #12
    Junior Member Renji Rodrigo's Avatar
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    This serie is not telescopic


     \sum^{\infty}_{k=1}\frac{1}{2k} -\frac{1}{2k+1}=\sum^{\infty}_{k=0}\frac{1}{2k+2} -\frac{1}{2k+3}=\sum^{\infty}_{k=0} \int^{1}_{0} x^{2k+1}(1-x)dx =


    =\int^{1}_{0} (1-x)x\sum^{\infty}_{k=0}  x^{2k}dx=\int^{1}_{0}\frac{x(1-x)}{1-x^2}dx = 1 - \ln (2)


    =\sum^{\infty}_{k=1}\frac{1}{(2k)(2k+1)} =  1 - \ln (2) .
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