Answer
Maclaurin series is: $\Sigma_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}$
and $R=\infty$
Work Step by Step
$f(x)=coshx=\Sigma_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}$
$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{\frac{x^{2n+2}}{(2n+2)!}}{\frac{x^{2n}}{(2n)!}}|$
$=\lim\limits_{n \to\infty}|\frac{x^{2}}{(2n+2)(2n+1)}|$
$=\lim\limits_{n \to\infty}|\frac{x^{2}}{\infty}|$
$=0\lt 1$
Therefore, the Maclaurin series converges for all values of $x$.
Maclaurin series is: $\Sigma_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}$
and $R=\infty$
