# Math Help - Infinite series (2)

1. ## Infinite series (2)

Prove that $\sum_{n=0}^{\infty} \frac{1}{(n!)^2} \notin \mathbb{Q}.$

2. 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

$\chi$ $\sigma$

3. Originally Posted by chisigma
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:

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

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

4. 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

$\chi$ $\sigma$

5. Originally Posted by chisigma
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. )