Answer
$$\sum^{\infty}_{n=1} (-1)^{n+1}(x-2)^n $$
$$\rho = 1$$
Work Step by Step
1. Find the Taylor series:
$$\frac{1}{1-x}= -1 + (\frac{1}{(-1)^2})(x-2) + (\frac{2}{(-1)^3})(x-2)^2\frac{1}{2!}+ (\frac{6}{(-1)^4})(x-2)^3\frac{1}{3!}+...
$$ $$
\frac{1}{1-x}= -1 + (x-2) - (x-2)^2 + (x-2)^3 ...
$$ $$
\frac{1}{1+x}= \sum^{\infty}_{n=1} (-1)^{n+1}(x-2)^n
$$
2. Use the ratio test to test for convergence:
$$\lim_{n \longrightarrow \infty}
\Bigg| \frac{ (-1)^{n+2}(x-2)^{n+1} }
{ (-1)^{n+1}(x-2)^n } \Bigg |
= \lim_{n \longrightarrow \infty} \Bigg| x\Bigg| = (-1)(x-2)$$
- Therefore, the series converges absolutely when $-x +2 \lt 1$.
$$\rho = 1$$
