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

There is an identity: $$\sum_{k=0}^n {{2k}\choose k} {{2n-2k}\choose{n-k}} = 4^n$$
From this, we can calculate, for all $x$ with $|x|<1/4$:
$$\left(\sum_{n=0}^\infty {{2n}\choose n} x^n\right)^2 = \sum_{n=0}^\infty \left( \sum_{k=0}^n {{2k}\choose k} {{2n-2k}\choose{n-k}}\right)x^n = \sum_{n=0}^\infty 4^n x^n = \frac{1}{1-4x}$$
We conclude:
$$\sum_{n=0}^\infty {{2n}\choose n} 4^{-n} = \lim_{x\to 1/4} \sum_{n=0}^\infty {{2n}\choose n} x^n = \lim_{x\to 1/4^-} \frac{1}{\sqrt{1-4x}} = \infty$$
