To add to what was written by Gerry Myerson, if $|w|<1$, we have by the usual power series expansion for $\log(1+w)$, $\log(1+w)=w-\frac{w^2}{2}+\frac{w^3}{3}-\frac{w^4}{4}+\cdots.\tag{$1$}$ Replacing $w$ by $-w$, we have $\log(1-w)=-w-\frac{w^2}{2}-\frac{w^3}{3}-\frac{w^4}{4}-\cdots.\tag{$2$}$ Subtracting $(2)$ from $(1)$, we obtain $\log(1+w)-\log(1-w)=\log\left(\frac{1+w}{1-w}\right)=2w+2\frac{w^3}{3}+2\frac{w^5}{5}+\cdots.\tag{$3$}$ For any $x>1$, there is a $w$ between $0$ and $1$ such that $x=\frac{1+w}{1-w}$. We can solve for $w$ explicitly in terms of $x$, obtaining $w=\frac{x-1}{x+1}$.
Thus we get an expansion for $\log x$ in terms of powers of $\frac{x-1}{x+1}$. This is not a power series expansion of $\ln x$, but it can be useful. For large $x$, the number $w$ is close to $1$, so the convergence is relatively slow. But for smallish $x$, there is usefully fast convergence. Euler, among others, used the series $(3)$ for computations.
Take for example $x=2$. The ordinary series for $\log(1+x)$ converges when $x=1$, but too slowly for practical use. However, in this case $w$ turns out to be $1/3$, and the series $(3)$ converges quite fast. For $x=3$, we get $w=1/2$, and again we get usefully fast convergence.