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Thread: series help

  1. #1
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    series help

    series help-series.png
    help i think that the n on the 2 is a misprint
    still help me to please
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  2. #2
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    Re: series help

    The $2^n$ in the denominator? The series wont converge without it so it looks fine to me.
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  3. #3
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    Re: series help

    Quote Originally Posted by bkbowser View Post
    The $2^n$ in the denominator? The series wont converge without it so it looks fine to me.
    What do you get as 1st 3 terms of this series?
    I'm a bit confused! Thanks.
    Trying to set up a looper...
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  4. #4
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    Re: series help

    Quote Originally Posted by DenisB View Post
    What do you get as 1st 3 terms of this series?
    I'm a bit confused! Thanks.
    Trying to set up a looper...
    Denis here is the general term.
    $\Large{{a_n} = \frac{{\prod\limits_{k = 1}^n {(2k - 1)} }}{{{2^n}\left[ {\prod\limits_{k = 1}^n {(3k - 2)} } \right]}}}$
    Thanks from topsquark and DenisB
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  5. #5
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    Re: series help

    How in hell did I write down 2^4 instead of 2^n !!
    Thanks Plato....
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  6. #6
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    Re: series help

    how do you solve it??
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  7. #7
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    Re: series help

    $\displaystyle \Large{{a_n} = \frac{{\prod\limits_{k = 1}^n {(2k - 1)} }}{{{\prod\limits_{k = 1}^n {(3k - 2)} } }}} <1$
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  8. #8
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    Re: series help

    u and a = numerator calculator
    v and b = denominator calculator

    Looper program:

    u=1:v=1:a=1:b=1
    loop n from 1 to ?
    a=a*u
    b=b*v
    t=a/(b*2^n) : current term value
    s=s+t : sum so far
    u=u+2
    v=v+3
    next n

    s = ~.79025 at n=15 and basically stays at that for eternity!!
    All kinda "silly" to me
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