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Math Help - Infinite Series

  1. #1
    Super Member Aryth's Avatar
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    Infinite Series

    Find the limit of the given sequence as n \to \infty.

    \frac{(n!)^2}{(2n)!}

    Now, I know that in the both the numerator and the denominator, a factor n! can be pulled out... From there, I have no idea where to go.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Aryth View Post
    Find the limit of the given sequence as n \to \infty.

    \frac{(n!)^2}{(2n)!}

    Now, I know that in the both the numerator and the denominator, a factor n! can be pulled out... From there, I have no idea where to go.
    No, you can't!

    (2n)!\neq 2!n!

    Let a_n=\frac{(n!)^2}{(2n)!}

    Then apply the ratio test:

    \lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\in  fty}\frac{\left((n+1)!\right)^2}{(2(n+1))!}\cdot\f  rac{(2n)!}{(n!)^2}=\lim_{n\to\infty}\frac{\left[\left(n+1\right)n!\right]^2}{(2n+2)(2n+1)(2n)!}\cdot\frac{(2n)!}{(n!)^2} =\lim_{n\to\infty}\frac{(n+1)^2}{(2n+2)(2n+1)}

    Can you take it from here?
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  3. #3
    Super Member Aryth's Avatar
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    Yes, I can take it from there. And I didn't mean to imply that

    (2n)! = 2!n!

    I actually didn't even know how to begin the problem, the only help I had received was that:

    (2n)! = 1(2)(3)...(n-2)(n-1)(n)...(2n) = n!(n+1)(n+2)...(2n)

    I didn't know what to do with it... And I'm glad I don't need it, what you helped me with makes much more sense.

    Thanks.
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  4. #4
    Super Member Random Variable's Avatar
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    Use Stirling's approximation.

    for large n  \frac{(n!)^2}{(2n)!} \approx \frac{(n^{n}e^{-n})^{2}}{(2n)^{2n}e^{-2n}} = \frac{n^{2n}e^{-2n}}{2^{2n}n^{2n}e^{-2n}} = \frac{1}{2^{2n}}

     \lim_{n \to \infty} \frac{1}{2^{2n}} = 0
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