Are these expressions (Gamma function and binomial) identical for $n\in \mathbb Z$

Question

For $n\in \mathbb Z$ and $n\ge 0$, prove that:

$$\frac{2\sqrt{\pi}\,\Gamma(\frac{1}{2}+n)}{\Gamma(n+1)}=\frac{\pi}{2^{2n-1}}\binom{2n}{n}$$

I started to prove.
We now that $\Gamma(\frac{1}{2})=\sqrt{\pi}$ so the left-hand-side will be:
$$\frac{2\Gamma(\frac{1}{2})\Gamma(\frac{1}{2}+n)}{\Gamma(1+n)}$$
which is equivalent to $2\beta(x,y)$ for $x=\frac{1}{2}$ and $y=n+\frac{1}{2}$
Also $\Gamma(n+1)=n!$ so the right-hand-side is:
$$\frac{\pi}{2^{2n-1}}\frac{(2n)!}{n!\,n!}=\frac{\pi}{2^{2n-1}}\frac{\Gamma(2n+1)}{\Gamma(n+1)\Gamma(n+1)}$$
But I seem to get no results!

Answer 1

Recall that: $$ \Gamma(n+1)= n!$$ $$ \Gamma(n+1)=n\Gamma(n)$$ $$ \Gamma(2n)= \frac{1}{\sqrt{2\pi}} 2^{2n - 1/2} \; \; \Gamma(n) \,\;\Gamma(n+\frac{1}{2})$$ So,if we consider the right hand side (RHS) : $$ \frac{\pi}{2^{2n-1}}\binom{2n}{n}=\frac{\pi}{2^{2n-1}} \frac{ (2n)!}{n! \, n! } = \frac{\pi}{2^{2n-1}} \frac{ 2n\Gamma(2n)}{\Gamma(n+1)\Gamma(n+1)}$$ using the last formula given above and applyinh it to $ \Gamma(2n) $: $$ RHS=\frac{\pi}{2^{2n-1}} \frac{ 2n}{\Gamma(n+1)\Gamma(n+1)} \frac{1}{\sqrt{2\pi}} 2^{2n - 1/2} \; \; \Gamma(n) \,\;\Gamma(n+\frac{1}{2}) $$ After cancellation

$$ RHS= \frac{2 \sqrt{\pi} \; \; n } {n \Gamma(n)\Gamma(n+1)} \Gamma(n) \,\;\Gamma(n+\frac{1}{2}) = LHS(\text{ Left Hand Side}) $$ Note: The last formula in the above recall is called the Duplicate formula.

Answer 2

Hint: Use the Gamma function duplication formula for $2n+1$: $$\Gamma(2n+1)=\frac{1}{\sqrt{2\pi}}2^{2n+1/2}\Gamma(n+\tfrac{1}{2})\Gamma(n+1)$$