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Math Help - factorials

  1. #1
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    factorials

    how do you get

    (2n-1)! / (2n+1)! to equal to

    (2n-1)! / (2n+1)(2n)(2n-1)!
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  2. #2
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    Quote Originally Posted by jeph View Post
    how do you get

    (2n-1)! / (2n+1)! to equal to

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

    WHAT DOES THIS HAVE TO DO WITH CALCULUS!!!

    (2n+1)! = (2n+1)(2n)(2n-1)(2n-2)....(2)(1)

    = (2n+1)(2n) * [ (2n-1)*(2n-2)*...*(2)(1) ]

    =(2n+1)(2n) * (2n-1)!
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  3. #3
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    well i thought it had something to do with calculus...its the first time I've seen the ! sign people used in a problem

    the question was:
    determine whether the sequence converges or diverges. If it converges, find the limit.

    i just needed to know how that step was done...
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jeph View Post
    well i thought it had something to do with calculus...its the first time I've seen the ! sign people used in a problem

    the question was:
    determine whether the sequence converges or diverges. If it converges, find the limit.

    i just needed to know how that step was done...
    TPH is right. just remember that the factorial of an integer is the product of all the integers from 1 up to that integer
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jeph View Post
    well i thought it had something to do with calculus...its the first time I've seen the ! sign people used in a problem

    the question was:
    determine whether the sequence converges or diverges. If it converges, find the limit.

    i just needed to know how that step was done...
    we're talking about a SEQUENCE and not a SERIES right. and i suppose you want the limit as n-->oo

    well in that case:
    yes, the sequence does converge--to zero

    lim{n-->oo}[(2n-1)!/(2n+1)! ] = lim{n-->oo}[(2n-1)!/(2n+1)(2n)(2n - 1)!]
    .........................................= lim{n-->oo}[1/(2n + 1)(2n)]
    .........................................= 0
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