Find $\displaystyle \sum\limits_{n\,=\,1}^{\infty }{\frac{\left( n! \right)^{2}}{(2n)!}}.$
This one is not hard.
hey these kinds I solve with Dalamber criterion.
taking the $\displaystyle \lim_{n\to\infty}{\frac{a_{n+1}}{a_{n}}}$ gives $\displaystyle \frac{1}{4}$ so it converges because $\displaystyle \frac{1}{4}<1$
but whatever, here integrals are required. So I'm out
Ok this one is killing me, I am making a mistake somewhere!
$\displaystyle \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}}=$$\displaystyle \frac{n\Gamma(n+1)\Gamma(n+1)}{\Gamma(2n+2)}=n\cdo t\text{B}(n+1,n+1)$
This gives us
$\displaystyle n\int_0^{1}t^n(1-t)^ndt$
so we have
$\displaystyle \sum_{n=1}^{\infty}n\int_0^{1}t^n(1-t)^ndt=\int_0^1\sum_{n=0}^{\infty}n(t-t^2)^ndt$
Now we know that $\displaystyle \forall{t}\in(0,1)\quad{t-t^2<1}$
So now we see that
$\displaystyle \frac{1}{1-(t-t^2)}=\sum_{n=0}^{\infty}(t-t^2)^n$
This implies that
$\displaystyle \frac{d}{dt}\bigg[\frac{1}{1-(t-t^2)}\bigg]=\sum_{n=1}^{\infty}n(1-2t)(t-t^2)^{n-1}$
So then
$\displaystyle (t-t^2)\cdot\frac{d}{dt}\bigg[\frac{1}{1-(t-t^2)}\bigg]=(1-2t)\sum_{n=0}^{\infty}n(t-t^2)^n$
and finally
$\displaystyle \frac{t-t^2}{1-2t}\cdot\frac{d}{dt}\bigg[\frac{1}{1-(t-t^2)}\bigg]=\sum_{n=0}^{\infty}n(t-t^2)^n$
Now the first part of that is equivalent to
$\displaystyle \frac{t-t^2}{(t^2-t+1)^2}$
So we have
$\displaystyle \sum_{n=1}^{\infty}\frac{(n!)^2}{(2n)!}=\int_0^{1} \frac{t-t^2}{(t^2-t+1)^2}dt$
Which I do not think is correct .
Where am I going wrong? Is this anywhere near correct?
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
$\displaystyle \frac{\sin^{-1}x}{\sqrt{1-x^{2}}} = \sum_{r=0}^{\infty} \frac{2^{2r} (r!)^{2}}{(2r+1)!}x^{2r+1}$
$\displaystyle \Rightarrow \frac{1}{1-x^{2}} + \frac{x \sin^{-1}x}{(1-x^{2})^{3/2}} = \sum_{r=0}^{\infty}\frac{2^{2r} (r!)^{2}}{(2r)!}x^{2r}$
Setting x = 1/2 gives the result:
$\displaystyle \boxed{\sum_{n=1}^{\infty} \frac{(n!)^{2}}{(2n)!} = \frac{2\pi\sqrt{3}}{27} + \frac{1}{3}}$
Mathstud: $\displaystyle \Gamma(2n+1) \cdot{n} \not= \Gamma(2n+2)$
Well sure, because you went straight from
$\displaystyle \int_{0}^{1} \frac{x}{(1-x+x^{2})^{2}} \; \mathrm{d}x$
to the answer!
Presumably it is not possible to find $\displaystyle \sum_{n=1}^{\infty} \frac{(n!)^{3}}{(3n)!}$?
Clearly, you can write each term as $\displaystyle (3n+1)(2n+1)\int^{1}_{0} x^{n}(1-x)^{2n}\;\mathrm{d}x \int^{1}_{0} x^{n}(1-x)^{n}\;\mathrm{d}x$, but that's not going to help.