Chapter 11 - Section 11.10 - Taylor and Maclaurin Series - 11.10 Exercises - Page 771: 23

$\Sigma_{n=0}^{\infty}\frac{(x-3)^{n}2^{n}e^{6}}{n!}$, $R=\infty$

$e^{2x}=\Sigma_{n=0}^{\infty}\frac{(x-3)^{n}2^{n}e^{6}}{n!}$ $\lim\limits_{n \to \infty}|\frac{\frac{(x-3)^{n+1}2^{n+1}}{(n+1)!}}{\frac{(x-3)^{n}2^{n}}{n!}}|$ $=|x-3|\lim\limits_{n \to\infty}|\frac{2}{n+1}|$ $=0\lt 1$ The radius of convergence is $R=\infty$.