Answer
$cosx=\Sigma_{n=0}^{\infty}(-1)^{n+1}\frac{(x-\frac{\pi}{2})^{2n+1}}{(2n+1)!}$
and
$R=\infty $
Work Step by Step
Taylor series centered at $a= \frac{\pi}{2}$ is
$cosx=\Sigma_{n=0}^{\infty}(-1)^{n+1}\frac{(x-\frac{\pi}{2})^{2n+1}}{(2n+1)!}$
$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{\frac{(-1)^{n+3}(x-\frac{\pi}{2})^{2(n+1)+1}}{(2(n+1)+1)!}}{\frac{(-1)^{n+1}(x-\frac{\pi}{2})^{2n+1}}{(2n+1)!}}|$
$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{\frac{(-1)^{n+3}(x-\frac{\pi}{2})^{2n+3}}{(2n+3)!}}{\frac{(-1)^{n+1}(x-\frac{\pi}{2})^{2n+1}}{(2n+1)!}}|$
$=\lim\limits_{n \to \infty}|\frac{(x-\frac{\pi}{2})^{2}}{(2n+2)(2n+3)}|$
$=0\lt 1$
This implies series converges for all the values of $x$ and radius of convergence is $R=\infty$.
