Answer
The Taylor series is $\Sigma_{n=0}^{\infty}\frac{(-1)^{n}}{3^{n}(n+1)}(x-4)^{n}$
and $R=3$
Work Step by Step
Use the given value of $f^{n}(4)$ in the formula for the Taylor series at $x=4$, then simplify
$\Sigma_{n=0}^{\infty}\frac{(-1)^{n}n!}{n!3^{n}(n+1)}(x-4)^{n}=\Sigma_{n=0}^{\infty}\frac{(-1)^{n}}{3^{n}(n+1)}(x-4)^{n}$
$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=|\frac{x-4}{3}|\lt 1$
Hence, the Taylor series is $\Sigma_{n=0}^{\infty}\frac{(-1)^{n}}{3^{n}(n+1)}(x-4)^{n}$
and $R=3$
