Question:

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

I got P_n = \sum_{k = 0}^n x^k for the nth Taylor polynomial, and 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 R_n < 10^{-6}. But I don't know what to put for \xi or x.