The gamma function is known to follow the below recurrence condition for all positive real numbers-
$$\Gamma\big(x+1\big)=x~\Gamma\big(x\big).~$$
Consider the function $\Gamma\big(n+\frac{1}{2}\big)$ where $n$ is a natural number. Using the recurrence described above, this could be broken down further as-
$$\Gamma\big(n+\frac{1}{2}\big)=(n-1)~\Gamma\big(n-1+\frac{1}{2}\big)=...=(n-1)!~\Gamma\big(1+\frac{1}{2}\big)=(n-1)!~\frac{\sqrt{\pi}}{2}$$
I was reading up online on this particular expansion, and saw formulae that didn't agree with the one that I've derived above. They state that-
$$\Gamma\big(n+\frac{1}{2}\big)=\frac{(2n)!}{2^{2n}n!}\sqrt{\pi}$$
What am I doing wrong in my derivation?
Observe the identity carefully- the correct version is this. $$\Gamma\big(n+\frac{1}{2}\big)=(n-1+\frac{1}{2})~\Gamma\big(n-1+\frac{1}{2}\big)$$