Answer
Maclaurin's series is: $\Sigma_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{(2n)!}$
and $R=\infty$
Work Step by Step
$f(x)=xcosx=\Sigma_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{2n!}$
$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{\frac{x^{2n+3}}{(2n+2)!}}{\frac{x^{2n+1}}{(2n)!}}|$
$=\lim\limits_{n \to\infty}|\frac{x^{2}}{(2n+2)(2n+1)}|$
$=\lim\limits_{n \to\infty}|\frac{x^{2}}{4n^{2}+6n+2}|$
$=0\lt 1$
Maclaurin's series is: $\Sigma_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{(2n)!}$
and $R=\infty$
