I am stuck simplifying the following expression: $$\frac{\left(\frac{k-1}2\right)!}{\left(\frac k2\right)!}$$ I knew the result is $\frac k{2(k+1)}$ or $\frac k{k+1}$ but I am unable to prove it. In the expansion of these factorials, there is a half-integer difference between the numerator's and denominator's expansions.
$\begingroup$
$\endgroup$
3
-
$\begingroup$ can you you tell me what you meant $$\frac{k!}{2}$$ or $$\frac{k}{2}!$$? $\endgroup$– Dr. Sonnhard GraubnerCommented Aug 27, 2016 at 11:14
-
1$\begingroup$ You must have copied the question wrong. That expression is never equal to $\frac k{2(k+1)}$ or $\frac k{k+1}$. Specifically, if $k$ is an integer, then one of $\frac{k-1}{2}$ and $\frac{k}{2}$ is not an integer; what does your factorial sign mean in this case? There is a way to define this, but it doesn't give those answers. $\endgroup$– TonyKCommented Aug 27, 2016 at 11:51
-
$\begingroup$ You can find some useful expressions here: mathworld.wolfram.com/GammaFunction.html $\endgroup$– N74Commented Aug 27, 2016 at 12:12
Add a comment
|