Non-equality with Gamma functions

Question

Let $n \in N$, $k \in Z_+$.

Show that $$ \frac{\Gamma\left(k+\frac 12\right)}{\Gamma\left(k+\frac 32\right)}\neq\frac{\Gamma\left(k+\frac{n-1}{2}\right)}{\Gamma\left(k+\frac{n-2}{2}\right)}\pi^{1-\frac n2}. $$

Thank you.

Answer 1

One of the first things on learns about $\Gamma$ function is its functional equation
$$z\,\Gamma(z)=\Gamma(z+1),\quad z\in\mathbb C$$ With it, your problem reduces to $$\frac{1}{k+\frac 12} \neq\frac{\Gamma\left(k+\frac{n-1}{2}\right)}{\Gamma\left(k+\frac{n-2}{2}\right)}\pi^{1-\frac n2}.$$ Also, multiplying both sides by $\Gamma\left(k+\frac{n-2}{2}\right)^2$ and using the duplication formula we get $$\frac{\Gamma\left(k+\frac{n-2}{2}\right)^2}{k+\frac 12} \neq 2^{1-(2k+n-2)} \sqrt{\pi} \Gamma(2k+n-2) \pi^{1-\frac n2}.$$ The squared $\Gamma$ on the left is a rational multiple of $\pi$. On the right there is a rational multiple of $\pi^{(3-n)/2}$. So, $n$ must be equal to $1$. Write out $\Gamma(k-1/2)^2$ and compare two sides.