$\frac{1}{4^n}\binom{1/2}{n} \stackrel{?}{=} \frac{1}{1+2n}\binom{n+1/2}{2n}$ - An identity for fractional binomial coefficients In trying to write an answer to this question:

calculate the roots of $z = 1 + z^{1/2}$ using Lagrange expansion

I have come across the identity
$$
\frac{1}{4^n}\binom{1/2}{n} = \frac{1}{1+2n}\binom{n+1/2}{2n}.
\tag{1}
$$
Could anyone help me prove this?  The few identities I know for binomial coefficients aren't enough to get anywhere useful and I don't see a way to account for the large difference in the number of factors in their respective numerators and denominators.

Simply compute ratios of each side. Introducing notation $\Theta_n f(n) = \frac{f(n+1)}{f(n)}$:
$$
   \Theta_n \frac{1}{4^n} \binom{1/2}{n} = \frac{1}{4} \Theta_n  \binom{1/2}{n} =\frac{1}{4} \frac{\Gamma(n+1) \Gamma\left(\frac{3}{2}-n\right)}{\Gamma(n+2) \Gamma\left(\frac{1}{2}-n\right)} = -\frac{2n-1}{8(n+1)}
$$
On another hand:
$$
 \Theta_n \frac{1}{2n+1} \binom{n+1/2}{2n} = \frac{2n+1}{2n+3} \Theta_n \binom{n+1/2}{2n} =  \frac{2n+1}{2n+3} \frac{\frac{\Gamma(n+5/2)}{\Gamma(2n+3) \Gamma(1/2-n)}}{\frac{\Gamma(n+3/2)}{\Gamma(2n+1) \Gamma(3/2-n)}} =  \frac{2n+1}{2n+3} \frac{(n+3/2)(1/2-n)}{(2n+2)(2n+1)} = -\frac{1}{8} \frac{2n-1}{n+1}
$$
The identity is obviously true for $n=0$, which finishes the proof.
