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Math Help - Today's calculation of integral #20

  1. #1
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    Today's calculation of integral #20

    Find \sum\limits_{n\,=\,1}^{\infty }{\frac{\left( n! \right)^{2}}{(2n)!}}.

    This one is not hard.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Krizalid View Post
    Find \sum\limits_{n\,=\,1}^{\infty }{\frac{\left( n! \right)^{2}}{(2n)!}}.

    This one is not hard.
    I don't want to type the whole thing unless I am on the right track (highlight)

    Transform into beta function
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  3. #3
    Math Engineering Student
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    I'm not gonna tell ya if that's the right track or not, that's something you have to find out.
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  4. #4
    Member javax's Avatar
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    hey these kinds I solve with Dalamber criterion.
    taking the \lim_{n\to\infty}{\frac{a_{n+1}}{a_{n}}} gives \frac{1}{4} so it converges because \frac{1}{4}<1
    but whatever, here integrals are required. So I'm out
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  5. #5
    Super Member wingless's Avatar
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    It's easy to solve this using Mathstud's approach.
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Krizalid View Post
    Find \sum\limits_{n\,=\,1}^{\infty }{\frac{\left( n! \right)^{2}}{(2n)!}}.

    This one is not hard.
    Ok this one is killing me, I am making a mistake somewhere!

    \frac{(n!)^2}{(2n)!}=\frac{\Gamma(n+1)\Gamma(n+1)}  {\Gamma(2n+1)}=\frac{n\Gamma(n+1)\Gamma(n+1)}{\Gam  ma(2n+1)\cdot{n}}= \frac{n\Gamma(n+1)\Gamma(n+1)}{\Gamma(2n+2)}=n\cdo  t\text{B}(n+1,n+1)

    This gives us

    n\int_0^{1}t^n(1-t)^ndt

    so we have

    \sum_{n=1}^{\infty}n\int_0^{1}t^n(1-t)^ndt=\int_0^1\sum_{n=0}^{\infty}n(t-t^2)^ndt

    Now we know that \forall{t}\in(0,1)\quad{t-t^2<1}

    So now we see that

    \frac{1}{1-(t-t^2)}=\sum_{n=0}^{\infty}(t-t^2)^n

    This implies that

    \frac{d}{dt}\bigg[\frac{1}{1-(t-t^2)}\bigg]=\sum_{n=1}^{\infty}n(1-2t)(t-t^2)^{n-1}

    So then

    (t-t^2)\cdot\frac{d}{dt}\bigg[\frac{1}{1-(t-t^2)}\bigg]=(1-2t)\sum_{n=0}^{\infty}n(t-t^2)^n

    and finally

    \frac{t-t^2}{1-2t}\cdot\frac{d}{dt}\bigg[\frac{1}{1-(t-t^2)}\bigg]=\sum_{n=0}^{\infty}n(t-t^2)^n

    Now the first part of that is equivalent to

    \frac{t-t^2}{(t^2-t+1)^2}

    So we have

    \sum_{n=1}^{\infty}\frac{(n!)^2}{(2n)!}=\int_0^{1}  \frac{t-t^2}{(t^2-t+1)^2}dt

    Which I do not think is correct .

    Where am I going wrong? Is this anywhere near correct?
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  7. #7
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    My initial thought was also to use the beta function, but it ends up being a bit messy. I think it's probably easier to use the formula

    \frac{\sin^{-1}x}{\sqrt{1-x^{2}}} = \sum_{r=0}^{\infty} \frac{2^{2r} (r!)^{2}}{(2r+1)!}x^{2r+1}

    \Rightarrow \frac{1}{1-x^{2}} + \frac{x \sin^{-1}x}{(1-x^{2})^{3/2}} = \sum_{r=0}^{\infty}\frac{2^{2r} (r!)^{2}}{(2r)!}x^{2r}

    Setting x = 1/2 gives the result:

    \boxed{\sum_{n=1}^{\infty} \frac{(n!)^{2}}{(2n)!} = \frac{2\pi\sqrt{3}}{27} + \frac{1}{3}}

    Mathstud: \Gamma(2n+1) \cdot{n} \not= \Gamma(2n+2)
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Dystopia View Post
    My initial thought was also to use the beta function, but it ends up being a bit messy. I think it's probably easier to use the formula

    \frac{\sin^{-1}x}{\sqrt{1-x^{2}}} = \sum_{r=0}^{\infty} \frac{2^{2r} (r!)^{2}}{(2r+1)!}x^{2r+1}

    \Rightarrow \frac{1}{1-x^{2}} + \frac{x \sin^{-1}x}{(1-x^{2})^{3/2}} = \sum_{r=0}^{\infty}\frac{2^{2r} (r!)^{2}}{(2r)!}x^{2r}

    Setting x = 1/2 gives:

    \sum_{n=1}^{\infty} \frac{(n!)^{2}}{(2n)!} = \frac{2\pi\sqrt{3}}{27} + \frac{1}{3}

    Mathstud: \Gamma(2n+1) \cdot{n} \not= \Gamma(2n+2)
    Dangit, stupid Gamma function.
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  9. #9
    Math Engineering Student
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    See here.

    It wasn't messy, besides Gamma function is not stupid.
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  10. #10
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    Quote Originally Posted by Krizalid View Post
    It wasn't messy
    Well sure, because you went straight from

    \int_{0}^{1} \frac{x}{(1-x+x^{2})^{2}} \; \mathrm{d}x

    to the answer!

    Presumably it is not possible to find \sum_{n=1}^{\infty} \frac{(n!)^{3}}{(3n)!}?

    Clearly, you can write each term as (3n+1)(2n+1)\int^{1}_{0} x^{n}(1-x)^{2n}\;\mathrm{d}x \int^{1}_{0} x^{n}(1-x)^{n}\;\mathrm{d}x, but that's not going to help.
    Last edited by Dystopia; June 29th 2008 at 04:31 PM.
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