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