Answer
$\infty$
Work Step by Step
$xe^{2x}=x\Sigma_{n=0}^\infty(2)^{n}\frac{x^{n}}{n!}=\Sigma_{n=0}^\infty(2)^{n}\frac{x^{n+1}}{n!}$
$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{(2)^{n+1}\frac{x^{n+2}}{n+1!}}{(2)^{n}\frac{x^{n+1}}{n!}}|$
$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{2x}{n+1}|$
$=0$
Thus, the series has radius of convergence is $\infty$ .
