I am trying to find the taylor series for $f(x)=(x-1)^3$ centered at $x=0$
Calculating derivative and evaluating them for $x=0$ we see that any derivative greater than and including the 5th derivative is 0.
So would the taylor series for this be,
$$\sum\frac{0x^2}{n!}=0$$
And then since the taylor series is $0$ when calculating the radius of convergence would it just be $\infty$ since when you use the ratio test for $\sum\frac{0x^2}{n!}=0$ you would get $lim=0$ therefore for all $x$ the series converges and the radius is $\infty$