1. ## Gamma Function

I have attached the question.

I have attempted this question from many angles yet I havn't been successful.
I would like some help on this.
Thank you

2. ## Re: Gamma Function

I tried proving it with induction.
First case: $\displaystyle n=0$ then we need to prove:
$\displaystyle \Gamma\left(\frac{3}{2}\right)=\frac{\sqrt{\pi}}{2 }$
This is true because
$\displaystyle \Gamma\left(\frac{3}{2}\right)=\Gamma\left(\frac{1 }{2}+1\right)$$\displaystyle =\frac{1}{2}\Gamma\left(\frac{1}{2}\right)$$\displaystyle =\frac{\sqrt{\pi}}{2}$

Suppose the statement is true for $\displaystyle n=k-1$ (induction hypothesis), we accept
$\displaystyle \Gamma\left(\frac{2k+1}{2}\right)$$\displaystyle =\frac{(2k-1)!}{2^{2k-1}(k-1)!}\sqrt{\pi} Proving the statement for \displaystyle n=k, we obtain: \displaystyle \Gamma\left(\frac{2k+3}{2}\right)$$\displaystyle =\Gamma\left(\frac{2k+1}{2}+1\right)=\left(\frac{2 k+1}{2}\right)\Gamma\left(\frac{2k+1}{2}\right)$
$\displaystyle =\left(\frac{2k+1}{2}\right)\frac{(2k-1)!}{2^{2k-1}(k-1)!}\sqrt{\pi}$
$\displaystyle =\frac{(2k+1)(2k-1)!}{2^{2k}(k-1)!}\sqrt{\pi}$
$\displaystyle =\frac{(2k+1)(2k)(2k-1)!}{2^{2k}(2k)(k-1)!}$ (add a factor $\displaystyle 2k$ in the numerator and the denominator)
$\displaystyle =\frac{(2k+1)!}{2^{2k+1}k!}\sqrt{\pi}$

This is what we wanted to prove.