Chapter 11 - Section 11.10 - Taylor and Maclaurin Series - 11.10 Exercises - Page 771: 25

$sin(\pi )=\Sigma_{n=0}^{\infty}\frac{(x-\pi)^{2n+1}(-1)^{n+1}}{(2n+1)!}$; The radius of convergence is $R=\infty$.

The 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 $R=\infty$.