I was reading a book in which the author claimed that using the fact that $\Gamma$ function is logarithmically convex, one can show that $\Gamma(1/2) = \sqrt \pi$
I already know that $\Gamma(n + 1/2) = \dfrac{(2n)!}{2^{2n}n!}\Gamma(1/2)$.
I feel that I have to use some convexity bounding, however all the calculation I tried failed.
Logarithmic convexity of $\Gamma$ implies $$\Gamma\left(\frac{x+y}2\right)\le\sqrt{\Gamma(x)\,\Gamma(y)}\tag{LC}.$$ Using (LC) with $x=n, y=n+1$ gives $$\Gamma(n+1/2)\le\sqrt{\Gamma(n)\,\Gamma(n+1)}=\frac{\Gamma(n+1)}{\sqrt{n}}.$$ Using (LC) with $x=n+1/2, y=n+3/2$ gives $$\Gamma(n+1)\le\sqrt{\Gamma(n+1/2)\,\Gamma(n+3/2)}=\Gamma(n+1/2)\,\sqrt{n+1/2}.$$ From the squeeze theorem, we see that $$\lim_{n\to\infty}\frac{\Gamma(n+1/2)\,\sqrt{n}}{\Gamma(n+1)}=1.\tag{LIMIT}$$ If you solve your formula $$\Gamma(n + 1/2) = \frac{(2n)!}{2^{2n}n!}\Gamma(1/2)$$ for $\Gamma(1/2)$ and let $n\to\infty$, considering (LIMIT) and https://en.wikipedia.org/wiki/Wallis_product, you'll arrive at the correct result, eventually.