I'm trying to find the radius of convergence of the series: $(-1)^{n+1} \frac{\bigl(\bigl|\tfrac{1-x}{1+x}\bigr|-1\bigl)^n} n$ representing the equation $\ln|(1-x)/(1+x)|$. Applying substitution to the Maclaurin series $\ln(1+x)$ gets me this. when I use the ratio test I'm given the inequality $(x<-3) \cup (-1/3<x)...$ what does this mean for a radius of convergence?
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by the root test
$\lim_\limits{n\to\infty}\left|\frac{\bigl(\bigl|\tfrac{1-x}{1+x}\bigr|-1\bigl)^n} n\right|^\frac 1n < 1\\ \lim_\limits{n\to\infty}n^\frac 1n|\left|\frac{1-x}{1+x}\right|-1| < 1$
$\lim_\limits{n\to\infty}n^\frac 1n = 1$
Which means that we only need to concern ourselves with:
$\left|\frac{1-x}{1+x}\right| - 1| < 1$
or
$\left| \frac{1-x}{1+x}\right| < 2$
Suppose $\frac{1-x}{1+x} > 0$ i.e. $-1 < x< 1$
$\frac{1-x}{1+x} < 2\\ -3x - 1 < 0\\ x > -\frac 13$
otherwise
$\frac{x-1}{1+x} < 2\\ \frac{-x-3}{1+x} < 0$
if $x \approx -3$ then $1+x < 0$
$-x-3 > 0\\ x < - 3$
We will converge when.
$x< - 3$ or $-\frac 13 < x < 1$
Now we should check the endpoints to see if these converge under closed (or half closed intervals)
If $|\frac {1-x}{1+x}|= 2, \sum_\limits{n=1}^\infty (-1)^n\frac{1^n} n$ converges
and when $x = 1, \sum_\limits{n=1}^\infty (0)^n\frac{1^n} n$ converges
$x\le - 3$ or $-\frac 13 \le x \le 1$