Answer
$$\sum^{\infty}_{n=1} x^n $$
$$\rho = 1$$
Work Step by Step
1. Find the Taylor series:
$$\frac{1}{1-x}= 1 + (\frac{1}{(1+0)^2})x + (\frac{2}{(1+0)^3})x^2\frac{1}{2!}+ (\frac{6}{(1+0)^4})x^3\frac{1}{3!}+...
$$ $$
\frac{1}{1-x}= 1 + x + x^2 + x^3 + x^4...
$$ $$
\frac{1}{1+x}= \sum^{\infty}_{n=1} x^n
$$
2. Use the ratio test to test for convergence:
$$\lim_{n \longrightarrow \infty}
\Bigg| \frac{ x^{n+1} }
{ x^n} \Bigg |
= \lim_{n \longrightarrow \infty} \Bigg| x\Bigg| = x$$
- Therefore, the series converges absolutely when $x \lt 1$.
$$\rho = 1$$
