(a) Determine the radius of convergence to the power series $f(x)= \displaystyle\sum\limits_{n=0}^\infty \frac{(2n)!}{(n!)^2}x^n$.
Should I use the ratio test?
(b) Assume the validity of the formula: $\displaystyle\int_{0}^{\frac{\pi}{2}}$sin$^{2n}(\theta) d\theta=$ $\displaystyle\frac{\pi(2n)!}{2^{2n+1}(n!)^2}$. for all $n\ge0$, and proof that $\displaystyle\int_{0}^{\frac{\pi}{2}}$ $\displaystyle\frac{1}{1-x\sin^{2}\theta}d\theta=$ $\displaystyle\frac{\pi}{2}f\left(\frac{x}{4}\right)$ for all $x\in \mathbb{R}$ with $\mid x \mid \lt 1 $, where $f$ describes the sum function from (a).
Should I look at geometric series?
(a) Since the terms in the power series are non-zero for $x\neq 0$, the ratio test is adapted.
(b) Yes: since $|x|\sin^2\theta\lt 1$ we can write $1/(1-x\sin^2\theta)$ as a power series and use the assumed result.