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