Finding the interval of convergence of $\sum^{\infty}_{n=0}\frac{(2n)!}{(n!)^2}x^n$ This is part of a question:

Use a similar trick to find nice upper and lower bounds for $\frac {(2n)!}{4^n(n!)^2}$, and thus finish finding the
  interval of convergence of $\sum^{\infty}_{n=0}\frac{(2n)!}{(n!)^2}x^n$.

Here is what "similar trick" refers to: $\int^{n}_1\log(x)dx<\log(n!)<\int^{n+1}_1\log(x)dx$.
The question requests to find the interval of convergence by finding the upper and lower bound of $\frac{(2n)!}{4^n(n!)^2}$. I do not know how to solve for it this way, but I know how to find the interval by using ratio test: $\lim_{n\to\infty}\frac{(2n+2)!|x|^{n+1}}{(n+1)!^2}/\frac{(2n)!|x|^n}{(n!)^2}=\lim_{n\to\infty}\frac{(2n+1)(2n+2)|x|}{(n+1)^2}=\lim_{n\to\infty}\frac{(4n+2)|x|}{n+1}=4|x|<1$, therefore the interval of convergence should be $-0.25<x<0.25$. 
Though this does not meet the requirement, at least I can tell if I do it right or wrong by checking the result. I then tried to utilize the inequality $$\int^{n}_1\log(x)dx<\log(n!)<\int^{n+1}_1\log(x)dx$$, and transform it into: $$n^{n}e^{1-n}<n!<(n+1)^{n+1}e^{-n}$$, then $$(2n)^{2n}e^{1-2n}<(2n)!<(2n+1)^{2n+1}e^{-2n}$$ and $$n^{2n}e^{2-2n}<(n!)^2<(n+1)^{2n+2}e^{-2n}$$.
Combining them, I get $\frac{(2n)^{2n}e}{(n+1)^{2n+2}}<\frac{(2n)!}{(n!)^2}<\frac{(2n+1)^{2n+1}}{n^{2n}e^2}$. Itried to compute the limits $\lim_{n\to\infty}\frac{(2n)^{2n}e}{(n+1)^{2n+2}}$ and $\lim_{n\to\infty}\frac{(2n+1)^{2n+1}}{n^{2n}e^2}$ but the limits are very complicated to solve for and I suspected that this was not the right way to do it. Does anyone know how to do it?
Sorry I misinterpreted the question. Let me rephrase it here: After finding the interior of the interval of convergence of the function here, to tell whether the function converges at $x=\pm 0.25$, we are required to produce an inequality to put the series $\frac {(2n)!}{4^n(n!)^2}$ within a certain range. I will revise the words as soon as possible.
 A: Let $$ a_n = \frac{(2n)!}{(n!)^2} x^n $$
Ratio test implies : $$
 \bigg|\frac{a_{n+1}}{a_n} \bigg| =
 \frac{(2n+2)(2n+1)}{(n+1)^2} |x| \rightarrow 4|x| <1 $$
So if $ |x|< \frac{1}{4}$, the series converges.
$ x= \frac{1}{4}$ : Wallis formula is $$ \lim \frac{(n!)^2 2^{2n}}{
(2n)!\sqrt{n}} =\sqrt{\pi } $$
Hence since $\sum \sqrt{\frac{\pi}{n} } =\infty$, it diverges.
$x=-\frac{1}{4}$ : It is alternating so that it converges.
So $\bigg[-\frac{1}{4},\frac{1}{4}\bigg)$
A: We have:
$$ C_n = \frac{(2n)!}{n!^2} = \binom{2n}{n}\tag{1}$$
and:
$$ \frac{C_{n+1}}{4 C_n}= \frac{2n+1}{2n+2}=1-\frac{1}{2n+2}\tag{2}$$
so it follows that:
$$ \left(\frac{C_{n+1}}{4 C_n}\right)^2 = 1-\frac{1}{n+1}+\frac{1}{4(n+1)^2} = \frac{n}{n+1}\left(1+\frac{1}{4n(n+1)}\right)\tag{3}$$
as well as:
$$ \frac{C_{N+1}}{2\cdot 4^N}= \prod_{n=1}^{N}\frac{C_{n+1}}{4C_n}=\sqrt{\prod_{n=1}^{N}\frac{n}{n+1}}\sqrt{\prod_{n=1}^{N}\left(1+\frac{1}{4n(n+1)}\right)}\tag{4}$$
so:
$$ \frac{2\,C_{N+1}\sqrt{N+1}}{4^{N+1}} = \prod_{n=1}^{N}\left(1+\frac{1}{4n(n+1)}\right)^{1/2}\tag{5}$$
but the RHS of $(5)$ is a convergent product for $N\to +\infty$ since:
$$ \sum_{n\geq 1}\frac{1}{4n(n+1)} = \frac{1}{4},\tag{6} $$
hence $(5)$ gives that $C_N$ behaves like:
$$ C_N \sim K \frac{4^N}{\sqrt{N}}\tag{7} $$
and that allows to find the radius of convergence of your series pretty easily. 
You may also prove that:
$$ \sum_{n\geq 0}\binom{2n}{n}x^n = \frac{1}{\sqrt{1-4x}}\tag{8} $$
holds for every $|x|<\frac{1}{4}$ by Vandermonde convolution or other techniques.
