Question:

Find the nth Taylor polynomial of $\displaystyle f(x) = (1-x)^{-1}$ at $\displaystyle x_0 = 0$. Find a value of n necessary for $\displaystyle P_n (x)$ to approximate $\displaystyle f(x)$ to within $\displaystyle 10^{-6}$ on $\displaystyle [0, 0.5]$

I got $\displaystyle P_n = \sum_{k = 0}^n x^k$ for the nth Taylor polynomial, and $\displaystyle R_n = \frac{f^{n+1}(\xi)}{(n+1)!} (x - x_0)^{n+1} = \frac{(n+1)! (1-\xi)^{-(n+2)}}{(n+1)!} x^{n+1} = \frac{1}{(1-\xi)^{n+2}} x^{n+1}$ for the remainder.

Not sure what to do next. I know I have to set up an inequality so that $\displaystyle R_n < 10^{-6}$. But I don't know what to put for $\displaystyle \xi$ or $\displaystyle x$.