Answer
Maclaurin series is: $\Sigma_{n=0}^{\infty}(-1)^{n}\frac{2^{n}x^{n}}{n!}$
and $R=\infty$
Work Step by Step
$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{\frac{2^{n+1}.x^{n+1}}{(n+1)!}}{\frac{2^{n}x^{n}}{n!}}|$
$=\lim\limits_{n \to\infty}|(\frac{2x}{n+1})|$
$=0 \lt 1$
Maclaurin series is: $\Sigma_{n=0}^{\infty}(-1)^{n}\frac{2^{n}x^{n}}{n!}$
and $R=\infty$
