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

As you noted, Stirling gives an estimate $4^{-n}\binom{2n}{n} \ge \frac{c}{\sqrt n}$, which shows divergence; as others have said, it's enough to show $4^{-n}\binom{2n}{n} \ge \frac cn$, which is much easier than the Stirling estimate.  Here are three more arguments for $\frac cn$.


*

*The most elementary one that I know (except perhaps the answer by user133281):
$$
4^n
= \left(\sum_{k=0}^n 1\cdot\binom nk\right)^2
\le \left(\sum_{k=0}^n 1^2\right) \left(\sum_{k=0}^n \binom nk^2\right)
= (n+1)\binom{2n}{n}
$$
That's Cauchy-Schwarz and the identity proved in this question.

*Using the beta function representation of binomial coefficients:
$$ \frac1{(2n+1)\binom{2n}{n}}
= \int_0^1 t^n (1-t)^n \,dt
\le \frac1{4^n} $$
since $\sqrt{t(1-t)}\le\frac12$ by AM/GM.

*Using the Wallis integral representation of central binomial coefficients, that is,
$$ \binom{2n}{n} = \frac{4^n}{2\pi} \int_0^{2\pi} \cos^{2n} t \,dt $$
(You can prove this by induction; a slicker way is to recall that, for $n\in\mathbb Z$,
$$\frac1{2\pi} \int_0^{2\pi} e^{int} \,dt = \begin{cases} 1 &\text{if $n=0$} \\ 0 &\text{otherwise} \end{cases}$$
Consequently, if $f(x) = \sum_{k=m}^n a_k x^k$ is a Laurent polynomial then $\frac1{2\pi} \int_0^{2\pi} f(e^{it}) \,dt = a_0$.  Now take $f(x)=(x+x^{-1})^{2n}$.)  Perhaps the simplest way to estimate the Wallis integral is
$$ \int_0^{2\pi} \cos^{2n} t \,dt
= 4\int_0^{\pi/2} \cos^{2n} t \,dt
\ge 4\int_0^{\pi/2} \cos^{2n} t \sin t \,dt
= \frac4{2n+1}
$$

