In obtaining a formula for the Catalan numbers I have got the expression $-\frac{1}{2}\binom{1/2}{n}(-4)^n$. All my efforts to show that this simplifies to $\frac{1}{n}\binom{2n-2}{n-1}$ have not succeeded. Is there some mistake in the original expression, and if not, how do I simplify it?
Thanks for your time.
As you have $n\in \mathbb{N}$, induction like in Nameless's answer seems to be the most natural way to prove your result.
However, here is a prove using properties of the $\Gamma$ function:
The binomial coefficient can be expressed in terms of the $\Gamma$ function as $$\binom{a}{b} = \frac{\Gamma(a+1)}{\Gamma(b+1) \Gamma(1+a-b)}.$$
Write $$C_1(n)=-\frac{1}{2}\binom{1/2}{n}(-4)^n = - \frac{(-4)^n \overbrace{\Gamma(3/2)}^{\sqrt\pi/2}}{2\Gamma(3/2-n) \Gamma(n+1)} = \frac{(-1)^{n+1} \sqrt\pi 4^{n-1}}{\Gamma(3/2-n) \Gamma(1+n)}$$ and [using the duplication formula $\Gamma(2z) =\Gamma(z) \Gamma(z+1/2)2^{2z-1}/\sqrt\pi$ with $z=n-1/2$] $$C_2(n)=\frac{1}{n}\binom{2n-2}{n-1} = \frac{\overbrace{\Gamma(2n-1)}^{\Gamma[2 (n-1/2)]}}{n \Gamma(n)^2} =\frac{4^{n-1}\Gamma(n-1/2)}{\sqrt\pi \,\Gamma(n+1)}.$$
Taking the ratio yields $$\frac{C_2(n)}{C_1(n)}=\frac{(-1)^{n+1}}{\pi} \Gamma(n-1/2) \Gamma(3/2-n).$$ Now use the reflection formula$$\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin(\pi z)}.$$ Setting $z=n-1/2$, we obtain $$\frac{C_2(n)}{C_1(n)}= \frac{(-1)^{n+1}}{\sin(\pi n -\pi/2)} = \frac{(-1)^n}{\cos(\pi n)} =1.$$
You want to show $$\frac{1}{n!}\frac{-1}2[\frac12(\frac12-1)...(\frac12-n+1)](-4)^n=\frac1n\frac{(2n-2)!}{(n-1)!(n-1)}\iff\\ \frac{-1}2[\frac12(\frac12-1)...(\frac12-n+1)](-4)^n=\frac{(2n-2)!}{(n-1)!}$$ Induction: It holds for $n=1$ and supposing it holds for $n$ we must prove $$\frac{-1}2[\frac12(\frac12-1)...(\frac12-n+1)(\frac12-n](-4)^{n+1}=\frac{(2n)!}{n!}$$ The left hand side is $$\frac{-1}2[\frac12(\frac12-1)...(\frac12-n+1)(\frac12-n](-4)^{n+1}= (\frac12-n)(-4)[\frac{-1}2[\frac12(\frac12-1)...(\frac12-n+1)](-4)^n]=(4n-2)\frac{(2n-2)!}{(n-1)!}$$ We must show $$\frac{(2n)!}{n!}=(4n-2)\frac{(2n-2)!}{(n-1)!}$$ which can easily be done
First we show that $$\binom{1/2}{n}={1\over n!}(1/2-1)(1/2-2)...(1/2-(n-1))=$$$$={1\over n!}(-1/2)(-3/2)...(-(2n-3)/2)=$$ $$={1\over n!}{(-1)^{n-1}\over2^n}1\cdot3\cdots(2n-3)\frac{2\cdot4\cdot6...(2n-4)}{2^{n-2}(n-2)!}=$$ $$={1\over n!}{(-1)^{n-1}\over2^{2n-2}}\frac{(2n-3)!}{(n-2)!}\frac{(2n-2)(2n-1)2n}{(n-1)n\cdot2^2\cdot(2n-1)}=$$ $$={1\over n!}{(-1)^{n-1}\over2^{2n}}\frac{(2n)!}{n!(2n-1)}=$$ $$={(-1)^{n-1}\over4^n(2n-1)}\frac{(2n)!}{n!n!}={(-1)^{n-1}\over4^n(2n-1)}\binom{2n}{n}$$ then we get $$\frac{-1}{2}\binom{1/2}{n}(-4)^n=\frac{-1}{2}{(-1)^{n-1}\over4^n(2n-1)}\binom{2n}{n}(-1)^n4^n=$$ $$={(-1)^{2n}\over2(2n-1)}\binom{2n}{n}={1\over2(2n-1)}{2n(2n-1)\over n(n-1)}\binom{2n-2}{n-2}=$$ $$={1\over n-1}\binom{2n-2}{n-2}={1\over n-1}\binom{2n-2}{n}$$
