Does the series $\sum_{n=1}^\infty\frac{(2n)!}{2^{2n}(n!)^2}$ converge or diverge. 
$$\sum_{n=1}^\infty\frac{(2n)!}{2^{2n}(n!)^2}$$

Can I have a hint for whether this series converges or diverges using the comparison tests (direct and limit) or the integral test or the ratio test?
I tried using the ratio test but it failed because I got 1 as the ratio. The integral test seems impossible to use here.
 A: For a more direct approach, you might directly expand the terms as follows:
$$\begin{align}
\frac{(2n)!}{4^n (n!)^2} &= \frac{1}{4^n}\frac{2n(2n-1)}{n^2}\frac{(2n-2)(2n-3)}{(n-1)^2}\cdots\frac{(4)(3)}{2^2} \frac{(2)(1)}{1^2} \\
&= \frac{2^n}{4^n}\frac{2n-1}{n} \frac{2n-3}{n-1}\cdots\frac{3}{2} \frac{1}{1} \\
&= \frac{4^n}{4^n} \frac{n - 1/2}{n} \frac{n - 3/2}{n-1} \cdots \frac{3/2}{2} \frac{1/2}{1}.
\end{align}$$
This is almost a telescoping product. By subtracting $1/2$ from each numerator (except the last), we get a smaller term that does telescope. Thus
$$
\frac{(2n)!}{4^n (n!)^2}
\geq
\frac{n-1}{n}\frac{n-2}{n-1} \cdots \frac{1}{2} \cdot (1/2) = \frac{1}{2n}.
$$
Thus the $n$th term of your series is bigger than $1/2n$, and diverges by comparison with the harmonic series
$$ \sum_{n \geq 1} \frac{1}{n}.$$
A: Every term of the series is
$$\frac{(2n)!}{2^{2n}(n!)^2}=\dfrac{\Gamma(2n+1)}{2^{2n}\Gamma^2(n)}=\dfrac{\Gamma(n+\frac12)}{n\Gamma(n)\sqrt{\pi}}>\dfrac{1}{2n}$$
by the formula 
$$\dfrac{\Gamma(n)}{\Gamma(2n)}=\dfrac{\sqrt{\pi}}{2^{2n-1}\Gamma(n+\frac12)}$$
A: HINT
We have that
$$\frac{(2n)!}{2^{2n}(n!)^2} =\frac1{4^n}\binom{2n}{n} \sim \frac{1}{\sqrt{2\pi n}}$$
indeed recall that by bounds and asymptotic formulas  for the binomial coefficient
$$\binom{2n}{n} \sim \frac{4^n}{\sqrt{2\pi n}}$$
then we can refer to limit comparison test.
Refer also to the related


*

*Show that that $\lim_{n\to\infty}\sqrt[n]{\binom{2n}{n}} = 4$
