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

Question

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.

Answer 1

This is your expression:

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