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Thread: Infinite series (2)

  1. #1
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    Infinite series (2)

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

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

    ... and...

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

    ... we obtain...

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

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

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

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

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
    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...

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

    ... and...

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

    ... we obtain...

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

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

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

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

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

    Kind regards

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

    Spoiler:

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

    but that is impossible because:

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

    $\displaystyle =\frac{(b+2)^2}{(b+1)^2((b+2)^2 -1)}=\frac{1}{(b+1)^2} + \frac{1}{(b+1)^2((b+2)^2 -1)} \leq \frac{9}{32} < 1.$
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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...

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

    After some effort I arrived to the identity...

    $\displaystyle \frac{1}{2\pi} \int_{-\pi}^{\pi} \cosh(2\cdot \sqrt{t}\cdot \cos x)\cdot dx = \sum_{n=0}^{\infty} \frac{t^{n}}{(n!)^{2}}$ (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 $\displaystyle |t|<1$, and that means that (2) is defined also for $\displaystyle -1 < t < 0$, where the argument of cosh (*) is imaginary...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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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...

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

    ... and...

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

    ... we obtain...

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

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

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

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

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
    Unfortunately, $\displaystyle \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 $\displaystyle \frac{1}{\pi} \int_{-1}^{1} \frac{\cosh 2\xi}{\sqrt{1-\xi^{2}}} d\xi = I_0(2) $, where $\displaystyle I_n(z) $ is the Modified Bessel Function of the First Kind

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