Convergence of $\sum \frac{(2n)!}{n!n!}\frac{1}{4^n}$ Does the series
$$\sum \frac{(2n)!}{n!n!}\frac{1}{4^n}$$
converges?
My attempt: Since the ratio test is inconclusive, my idea is to use the Stirling Approximation for n!
$$\frac{(2n)!}{n!n!4^n} \sim (\frac{1}{4^n} \frac{\sqrt{4\pi n}(\frac{2n}{e})^{2n}}{\sqrt{2 n \pi} \sqrt{2n \pi} (\frac{n}{e})^{2n}} =\frac{(2)^{2n}}{4^n \sqrt{n \pi}}$$
The series of the secomd term diverges. It is correct to conclude thatthe series diverges?
Another ideas are welcome!
Thanks

A much simpler way:
$$
\frac{a_{n+1}}{a_n}=\frac{(2n+2)(2n+1)}{4(n+1)(n+1)}=\frac{2n+1}{2n+2}\ge
\sqrt{\frac{n}{n+1}},
$$
since
$$
\left(\frac{2n+1}{2n+2}\right)^2=\left(1-\frac{1}{2n+2}\right)^2\ge 1-\frac{2}{2n+2}
=\frac{n}{n+1},
$$
and hence
$$
a_n=a_1\prod_{k=2}^n\frac{a_{k}}{a_{k-1}}\ge a_1\prod_{k=2}^n\sqrt{\frac{k-1}{k}}
=\frac{a_1}{\sqrt{n}},
$$
and hence
$$
\sum_{n=1}^\infty a_n=\infty.
$$
Note that
$$
\frac{a_{n+1}}{a_n}=\frac{(2n+2)(2n+1)}{4(n+1)(n+1)}=\frac{2n+1}{2n+2}=1-\frac{1}{2n+2}=1-\frac{1}{2n}+{\mathcal O}\left(\frac{1}{n^2}\right).
$$
This series diverges due to Gauss Test otherwise known as Raabe's Test.
