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