# Today's calculation of integral #20

• Jun 29th 2008, 11:31 AM
Krizalid
Today's calculation of integral #20
Find $\sum\limits_{n\,=\,1}^{\infty }{\frac{\left( n! \right)^{2}}{(2n)!}}.$

This one is not hard. :)
• Jun 29th 2008, 12:01 PM
Mathstud28
Quote:

Originally Posted by Krizalid
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
• Jun 29th 2008, 12:03 PM
Krizalid
I'm not gonna tell ya if that's the right track or not, that's something you have to find out.
• Jun 29th 2008, 12:36 PM
javax
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(Shake)
• Jun 29th 2008, 12:40 PM
wingless
It's easy to solve this using Mathstud's approach.
• Jun 29th 2008, 01:34 PM
Mathstud28
Quote:

Originally Posted by Krizalid
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 (Angry).

Where am I going wrong? Is this anywhere near correct?
• Jun 29th 2008, 01:57 PM
Dystopia
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: $\Gamma(2n+1) \cdot{n} \not= \Gamma(2n+2)$
• Jun 29th 2008, 01:59 PM
Mathstud28
Quote:

Originally Posted by Dystopia
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. (Blush)
• Jun 29th 2008, 02:52 PM
Krizalid
See here.

It wasn't messy, besides Gamma function is not stupid. (Sun)
• Jun 29th 2008, 03:05 PM
Dystopia
Quote:

Originally Posted by Krizalid
It wasn't messy

Well sure, because you went straight from

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

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.