## nth Taylor Polynomial to approximate f(x) within 10^-6

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 $x$.