# Series expansion for $\frac{1}{x+1}$ in terms of $\frac{1}{x-1}$

I would like to expand a function as $$\frac{1}{x+1} = \frac{1}{x-1+2} = \frac{1}{x-1} \frac{1}{ 1+\frac{2}{x-1}} = \frac{1}{x-1} \left[ 1- \frac{2}{x-1} + \left(\frac{2}{x-1}\right)^2 + \cdots \right]$$

I tried

Series[1/(x + 1), {x, Infinity + 1, 3}]

apprently it does not work. Is there any robust way to realize this kind of expansion? Assume $|x|$ is sufficiently large.

• I don't know if this answers your question but here is a workaround: Series[Unevaluated[1/((x - 1) + 2)] /. HoldPattern[x - 1] -> t, {t, Infinity, 3}] /. t -> (x - 1)} – John Aug 2 '14 at 15:20
• Thanks. I managed to get a solution, 1/(x + 1) /. x -> t + 1; Series[%, {t, Infinity, 3}]; % /. t -> x - 1 – Guest25874 Aug 2 '14 at 15:23

expr1 = 1/(x + 1);


This is the the change of variables:

 sl = Solve[x - 1 == z, x][[1, 1]]

(*  x -> 1 + z  *)


Substituted to the expression it yields this:

expr2 = expr1 /. sl

(*  1/(2 + z)   *)


Now it can be expaneded as you need:

Series[expr1 /. sl, {z, Infinity, 5}] /. z -> HoldForm[x - 1]

(*
SeriesData[
HoldForm[x - 1],
DirectedInfinity[1], {1, -2, 4, -8, 16}, 1, 6, 1]   *)


Just try