Proving the Induction Step for the Gamma Function Identity

[Dustinsfl]

\Gamma \left(n+\frac{1}{2}\right) = \frac{(2n)!\sqrt{\pi}}{n!2^{2n}}

P(0) is true.

Assume true for a fixed by arbitrary integer k\geq n.

P(k + 1) = \Gamma \left(k+1+\frac{1}{2}\right) = \frac{\left[2\left(k+1\right)\right]!\sqrt{\pi}}{(k+1)!2^{2(k+1)}}

\Gamma \left(k+\frac{1}{2}\right)\left(k+\frac{1}{2}\right) = \frac{(2k)!\sqrt{\pi}}{k!2^{2k}}\left(k+\frac{1}{2 }\right)

\Rightarrow \Gamma \left(k+1+\frac{1}{2}\right) = \frac{(2k)!(2k+1)\sqrt{\pi}}{k!2^{2k}2}

\Rightarrow \Gamma \left(k+1+\frac{1}{2}\right) = \frac{(2k)!(2k+1)\sqrt{\pi}}{k!2^{2k+1}}

Not sure what to do now.

 

[tiny-tim]

Hi Dustinsfl!

Fine so far.

Now try multiplying the top and bottom by (2k + 2) ​

 

[Dustinsfl]

Thanks