Generating function of gamma quotient.

Question

I would like to show that the following is true. Let $g=\sum_i a_i x^i$ with $$ a_i=C\frac{\Gamma (i+0.5)}{\Gamma(i+2)} $$ and normalisation constant obtained from $\sum_i a_i=1$. Given this we should have $$ g=\frac 1{1+\sqrt{1-x}} $$ I have no intuition on how to approach this and would welcome any hints or advice.

Answer 1

\begin{align} \sum_{n=0}^{\infty}\frac{\Gamma(n+1/2)}{\Gamma(n+2)}x^n &=\frac{1}{\Gamma(3/2)}\sum_{n=0}^{\infty}x^n\mathrm{B}\left(n+\frac12,\frac32\right) \\\color{gray}{[\text{integral representation of }\mathrm{B}]}\quad &=\frac{2}{\sqrt\pi}\sum_{n=0}^{\infty}x^n\int_0^1y^{n-1/2}(1-y)^{1/2}~dy \\\color{gray}{[\textstyle\sum\leftrightarrow\int\text{ and geometric series}]}\quad &=\frac{2}{\sqrt\pi}\int_0^1\sqrt\frac{1-y}{y}\frac{dy}{1-xy} \\\color{gray}{[\text{substitution }\sqrt{(1-y)/y}=z]}\quad &=\frac{4}{\sqrt\pi}\int_0^\infty\frac{z^2~dz}{(1+z^2)(1-x+z^2)} \\\color{gray}{[\text{partial fractions}]}\quad &=\frac{4}{x\sqrt\pi}\int_0^\infty\left(\frac{1}{1+z^2}-\frac{1-x}{1-x+z^2} \right)dz \\\color{gray}{[\text{elementary integrals}]}\quad &=\frac{2\sqrt\pi}{x}(1-\sqrt{1-x})=\color{blue}{\frac{2\sqrt\pi}{1+\sqrt{1-x}}}. \end{align}