# Thread: Today's calculation of integral #20

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

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

This one is not hard.

2. 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

3. I'm not gonna tell ya if that's the right track or not, that's something you have to find out.

4. 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

5. It's easy to solve this using Mathstud's approach.

6. 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 .

Where am I going wrong? Is this anywhere near correct?

7. 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: $\Gamma(2n+1) \cdot{n} \not= \Gamma(2n+2)$

8. 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.

9. See here.

It wasn't messy, besides Gamma function is not stupid.

10. 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.