Q:

Asymptotic expansion of $\ln\left(\frac{x+a}{x-a}\right)$ in form of $\sum\limits_{n=0}^\infty a_n \left(\frac{1}{x}\right)^n$?

How can I find an expansion for $f(x)=\ln\left(\dfrac{x+a}{x-a}\right)$ in terms of powers of $x$, in the form of:
$$f(x)=\sum_{n=0}^\infty a_n \left(\frac{1}{x}\right)^n$$
When I try a Taylor expansion, I always end up with a very complicated (double) series in terms of $x^n$ for $n\geq0$.

A:

For the sake of having an answer: consider that
$$
f(x)=\ln\left(1+\frac{a}{x}\right)-\ln\left(1-\frac{a}{x}\right).
$$

