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Math Help - Infinite series (2)

  1. #1
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    Infinite series (2)

    Prove that \sum_{n=0}^{\infty} \frac{1}{(n!)^2} \notin \mathbb{Q}.
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  2. #2
    MHF Contributor chisigma's Avatar
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    Combining toghether the well known expressions...

    \int_{-\pi}^{\pi} \cos^{2n} x\cdot dx = \frac{2\cdot \pi}{2^{2n}}\binom {2n}{n} = \frac{2\cdot \pi}{2^{2n}}\cdot \frac{(2n)!}{(n!)^{2}} (1)

    ... and...

    \cosh x = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!} (2)

    ... we obtain...

    \int_{-\pi}^{\pi} \cosh(2\cdot \sqrt{t}\cdot \cos x)\cdot dx= \int_{-\pi}^{\pi} \sum_{n=0}^{\infty} \frac{2^{2n}\cdot t^{n}\cdot \cos^{2n} x}{(2n)!} \cdot dx =

    = 2 \pi \cdot \sum_{n=0}^{\infty} \frac{2^{2n}\cdot t^{n}}{(2n)!} \cdot \int_{-\pi}^{\pi} \cos^{2n} x\cdot dx= 2 \pi \cdot \sum_{n=0}^{\infty} \frac{t^{n}}{(n!)^{2}} (3)

    Setting t=1 in (3) we arrive to see that is...

     \sum_{n=0}^{\infty} \frac{1}{(n!)^{2}} = \frac{1}{2 \pi} \int_{-\pi}^{\pi} \cosh(2\cdot \cos x)\cdot dx = \frac{1}{\pi} \int_{-1}^{1} \frac{\cosh 2\xi}{\sqrt{1-\xi^{2}}} \cdot d\xi (4)

    The integral in (4) can probably be 'attacked' using the residue theorem... if we demonstrate that it does'nt contain the term \pi the goal is realized...

    Kind regards

    \chi \sigma
    Last edited by chisigma; May 23rd 2009 at 11:05 AM. Reason: an x instead of a t in (3)... sorry...
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  3. #3
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    Quote Originally Posted by chisigma View Post
    Combining toghether the well known expressions...

    \int_{-\pi}^{\pi} \cos^{2n} x\cdot dx = \frac{2\cdot \pi}{2^{2n}}\binom {2n}{n} = \frac{2\cdot \pi}{2^{2n}}\cdot \frac{(2n)!}{(n!)^{2}} (1)

    ... and...

    \cosh x = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!} (2)

    ... we obtain...

    \int_{-\pi}^{\pi} \cosh(2\cdot \sqrt{t}\cdot \cos x)\cdot dx= \int_{-\pi}^{\pi} \sum_{n=0}^{\infty} \frac{2^{2n}\cdot t^{n}\cdot \cos^{2n} x}{(2n)!} \cdot dx =

    = 2 \pi \cdot \sum_{n=0}^{\infty} \frac{2^{2n}\cdot t^{n}}{(2n)!} \cdot \int_{-\pi}^{\pi} \cos^{2n} x\cdot dx= 2 \pi \cdot \sum_{n=0}^{\infty} \frac{t^{n}}{(n!)^{2}} (3)

    Setting t=1 in (3) we arrive to see that is...

     \sum_{n=0}^{\infty} \frac{1}{(n!)^{2}} = \frac{1}{2 \pi} \int_{-\pi}^{\pi} \cosh(2\cdot \cos x)\cdot dx = \frac{1}{\pi} \int_{-1}^{1} \frac{\cosh 2\xi}{\sqrt{1-\xi^{2}}} \cdot d\xi (4)

    The integral in (4) can probably be 'attacked' using the residue theorem... if we demonstrate that it does'nt contain the term \pi the goal is realized...

    Kind regards

    \chi \sigma
    the problem is not that complicated! the idea is the same as the proof of irrationality of e. suppose \frac{a}{b}=\sum_{n=0}^{\infty} \frac{1}{(n!)^2}, for some positive integers a,b. then:

    Spoiler:

    ab(b-1)!^2 = \frac{(b!)^2 a}{b}=(b!)^2 \sum_{n=0}^b \frac{1}{(n!)^2} + (b!)^2\sum_{n=b+1}^{\infty} \frac{1}{(n!)^2}. therefore, since (b!)^2 \sum_{n=0}^b \frac{1}{(n!)^2} is an integer, A_n=(b!)^2\sum_{n=b+1}^{\infty} \frac{1}{(n!)^2} must be an integer too.

    but that is impossible because:

    0 < A_n=\sum_{n=b+1}^{\infty} \frac{1}{[n(n-1) \cdots (b+1)]^2} \leq \sum_{n=b+1}^{\infty} \frac{1}{(b+1)^2(b+2)^{2n-2b-2}}

    =\frac{(b+2)^2}{(b+1)^2((b+2)^2 -1)}=\frac{1}{(b+1)^2} + \frac{1}{(b+1)^2((b+2)^2 -1)} \leq \frac{9}{32} < 1.
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  4. #4
    MHF Contributor chisigma's Avatar
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    Some years ago in an other 'matemathical challenge' I had to find the 'generating function' of the sequence...

    a_{n} = \frac{1}{(n!)^{2}} (1)

    After some effort I arrived to the identity...

    \frac{1}{2\pi} \int_{-\pi}^{\pi} \cosh(2\cdot \sqrt{t}\cdot \cos x)\cdot dx = \sum_{n=0}^{\infty} \frac{t^{n}}{(n!)^{2}} (2)

    I don't think that (2) in very important for some application, but in any case it is interesting. As most of series of this type in converges for |t|<1, and that means that (2) is defined also for -1 < t < 0, where the argument of cosh (*) is imaginary...

    Kind regards

    \chi \sigma
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  5. #5
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by chisigma View Post
    Combining toghether the well known expressions...

    \int_{-\pi}^{\pi} \cos^{2n} x\cdot dx = \frac{2\cdot \pi}{2^{2n}}\binom {2n}{n} = \frac{2\cdot \pi}{2^{2n}}\cdot \frac{(2n)!}{(n!)^{2}} (1)

    ... and...

    \cosh x = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!} (2)

    ... we obtain...

    \int_{-\pi}^{\pi} \cosh(2\cdot \sqrt{t}\cdot \cos x)\cdot dx= \int_{-\pi}^{\pi} \sum_{n=0}^{\infty} \frac{2^{2n}\cdot t^{n}\cdot \cos^{2n} x}{(2n)!} \cdot dx =

    = 2 \pi \cdot \sum_{n=0}^{\infty} \frac{2^{2n}\cdot t^{n}}{(2n)!} \cdot \int_{-\pi}^{\pi} \cos^{2n} x\cdot dx= 2 \pi \cdot \sum_{n=0}^{\infty} \frac{t^{n}}{(n!)^{2}} (3)

    Setting t=1 in (3) we arrive to see that is...

     \sum_{n=0}^{\infty} \frac{1}{(n!)^{2}} = \frac{1}{2 \pi} \int_{-\pi}^{\pi} \cosh(2\cdot \cos x)\cdot dx = \frac{1}{\pi} \int_{-1}^{1} \frac{\cosh 2\xi}{\sqrt{1-\xi^{2}}} \cdot d\xi (4)

    The integral in (4) can probably be 'attacked' using the residue theorem... if we demonstrate that it does'nt contain the term \pi the goal is realized...

    Kind regards

    \chi \sigma
    Unfortunately,  \frac{1}{\pi} \int_{-1}^{1} \frac{\cosh 2\xi}{\sqrt{1-\xi^{2}}} d\xi is not very nice to compute.

    It turns out that  \frac{1}{\pi} \int_{-1}^{1} \frac{\cosh 2\xi}{\sqrt{1-\xi^{2}}} d\xi = I_0(2) , where  I_n(z) is the Modified Bessel Function of the First Kind

    If you look at the identities for  I_0(z) , it becomes evident that showing the irrationality of  I_0(2) is a fairly difficult thing to do.
    (Also, check out the last identity listed in the link. )
    Last edited by chiph588@; March 28th 2010 at 11:47 AM.
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