Answer
$f(x)=\Sigma_{n=0}^{\infty}(n+1)x^{n}$
and
$R=1$
Work Step by Step
Given: $f^{n}(0)=(n+1)!$ for $n=0,1,2,...$
Maclaurin series: $f(x)=f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}(x)^{2}+...$
$f(x)=1!+\frac{2!}{1!}x+\frac{3!}{2!}x^{2}+\frac{4!}{3!}x^{2}+...$
$=1+2x+3x^{2}+4x^{3}+...$
From this pattern we get
$f(x)=\Sigma_{n=0}^{\infty}(n+1)x^{n}$
Use the ratio test:
$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{(n+2)x^{n+1}}{(n+1)x^{n}}|$
$=\lim\limits_{n \to \infty}|\frac{(1+2/n)x}{(1+1/n)}|$
$=|x|$
This series converges when $|x| \lt 1$
Thus, the series has radius of convergence is $1$ .
