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 $n\geq0$.

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    $\begingroup$ Hint: $$f(x)=\ln\left(1+\frac{a}{x}\right)-\ln\left(1-\frac{a}{x}\right)$$ $\endgroup$ – Start wearing purple Aug 2 '13 at 8:55

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

  • $\begingroup$ Why is this community wiki ? $\endgroup$ – mick Oct 22 '14 at 19:59
  • $\begingroup$ @mick Because this was already posted (as a comment) by another user hence (I decided that) rep points would be undue. $\endgroup$ – Did Oct 22 '14 at 20:12
  • $\begingroup$ Such high morals :) thanks @Did $\endgroup$ – mick Oct 22 '14 at 20:23

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