 Solution

Set $h=x+3$, so $x=-3(1-h/3)$. The negative binomial series gives $(1-u{)}^{-4}={\sum }_{n=0}^{\infty }(\genfrac{}{}{0ex}{}{n+3}{3})u^n$ for $|u|<1$. Hence the exact Taylor series isThe initial terms are $\frac{7}{81}[1+4h/3+10h^2/9+20h^3/27+\cdots ]$. Equivalently the coefficient of $(x+3{)}^n$ is $7(n+3)!/[6n!3^{n+4}]$, obtained by differentiating $7x^{-4}$ repeatedly and evaluating at $-3$.

The radius of convergence is three, set by the pole at $x=0$. At either endpoint $h=\pm 3$, the terms fail to tend to zero, so neither endpoint is included.