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