How to simplify the Gamma functions such that the Binomial Theorem can be used?

Question

I have the following expression: \begin{align} \sum^\infty_{k=0} \frac{\Gamma\left( \frac{2k+1}{b} \right) }{\Gamma(2k+1)} (-1)^ky^k. \end{align} Is there any way to manipulate this expression to equate to something like $(1+y)^{-n}$?

What I know so far is

1. the factorials can be expressed in terms of Gamma functions, i.e. $n!=\Gamma(n+1)$;
2. The Binomial Theorem: $$ (1+x)^{-n} = \sum^\infty_{k=0} {{n+k-1}\choose{k}} (-1)^k x^k, $$ since ${{-n}\choose{k}}= (-1)^k{{n+k-1}\choose{k}}$.