Prove that, for positive integral n:

$\mathbf{\Gamma }\left(n+\frac{1}{2}\right)\mathbf{=}\frac{\mathbf{1}\mathbf{·}\mathbf{3}\mathbf{·}\mathbf{5}\mathbf{\cdots }\mathbf{(}\mathbf{2}\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{)}}{{\mathbf{2}}^{\mathbf{n}}}\sqrt{\mathbf{\pi }}\mathbf{=}\frac{\mathbf{\left(}\mathbf{2}\mathbf{n}\mathbf{\right)}\mathbf{!}}{{\mathbf{4}}^{\mathbf{n}}\mathbf{n}\mathbf{!}}\sqrt{\mathbf{\pi }}$

The value of a common Gamma function of $\frac{1}{2}$.

$\Gamma \left(\frac{1}{2}\right)=\sqrt{\pi }$

The answer can be found out using mathematical induction. For ,n=1 calculate the Gamma function

$\begin{array}{c}\mathrm{\Gamma }\left(\frac{3}{2}\right)=\mathrm{\Gamma }\left(1+\frac{1}{2}\right)\\ \mathrm{\Gamma }\left(1+\frac{1}{2}\right)=\frac{1}{2^1}\sqrt{\pi }\\ =\frac{\sqrt{\pi }}{2}\end{array}$

Prove that the solution exists for $n+1$.

$\begin{array}{c}\mathrm{\Gamma }\left(n+1+\frac{1}{2}\right)=\frac{1·3·5\cdots (2n-1)·\left(2\right(n+1)-1)}{2^{n+1}}\sqrt{\pi }\\ =\frac{\left(2\right(n+1\left)\right)!}{4^{n+1}(n+1)!}\sqrt{\pi }\end{array}$

Continue the equation as:

role="math" localid="1664341142875" $\begin{array}{c}\frac{1·3·5\cdots (2n-1)·(2n+1)}{2^{n}2}=\frac{\left(2\right(n+1\left)\right)!}{4^{n+1}(n+1)!}\\ \frac{1·3·5\cdots (2n-1)}{2^n}\frac{2n+1}{2}=\frac{(2n+2)!}{4^{n+1}(n+1)!}\end{array}$

In the above equation, use the assumption of mathematical induction for the first fraction on the left hand side.

$\frac{\left(2n\right)!}{4^{n}n!}\frac{2n+1}{2}=\frac{(2n+1)!}{2·4^{n}n!}$

$\frac{(2n+2)!}{4^{n+1}(n+1)!}$

Continue the simplification.

$\begin{array}{c}\frac{(2n+2)!}{4^{n+1}(n+1)!}=\frac{(2n+2)(2n+1)!}{{44}^n(n+1)n!}\\ \frac{(2n+1)!}{2·4^{n}n!}=\frac{(2n+2)(2n+1)!}{{44}^n(n+1)n!}\end{array}$

On dividing these equations by a common term, the answer comes to .

The proof is complete