Is this a valid proof that absolute value of $\ln(1-x) > \ln(1+x)$ for $0 < x< 1$?

Question

I used CodeCogs to create the equations in the proof. Is there any way to directly use asciimath or something else?

Answer 1

Here is a proof with Taylor expansion for $|\ln(1-x)| > |\ln(1+x)|$ given $0<x<1$:

\begin{align}\ln(1-x)=-\sum_{n=1}^{\infty}\frac{x^{n}}{n}, \ln(1+x)=-\sum_{n=1}^{\infty} (-1)^{n}\frac{x^n}{n}\end{align}

\begin{align}|\ln(1+x)|=|\sum_{n=1}^{\infty} (-1)^{n}\frac{x^n}{n}|<\sum_{n=1}^{\infty} |\frac{x^n}{n}|=|\sum_{n=1}^{\infty}\frac{x^n}{n}|\end{align}

\begin{align}|\frac{\ln(1-x)}{\ln(1+x)}|=\frac{|-\sum_{n=1}^{\infty}\frac{x^{n}}{n}|}{|\sum_{n=1}^{\infty} (-1)^{n}\frac{x^n}{n}|}>\frac{|\sum_{n=1}^{\infty}\frac{x^{n}}{n}|}{|\sum_{n=1}^{\infty} \frac{x^n}{n}|}=1\end{align}\begin{align} |\ln(1-x)| > |\ln(1+x)|\end{align}