Answer
$sin(\pi )=\Sigma_{n=0}^{\infty}\frac{(x-\pi)^{2n+1}(-1)^{n+1}}{(2n+1)!}$
The radius of convergence is $R=\infty$.
Work Step by Step
Taylor series centered at $a= \pi$ is
$-(x-\pi)+\frac{(x-\pi)^{3}}{3!}-\frac{(x-\pi)^{5}}{5!}+...$
$sin(\pi )=\Sigma_{n=0}^{\infty}\frac{(x-\pi)^{2n+1}(-1)^{n+1}}{(2n+1)!}$
The radius of convergence is $R=\infty$.