Series expansion for $\frac{x}{1- \frac{1}{x}}$

Question

I would like to expand $\frac{x}{1- \frac{1}{x}}$ as $$\frac{x}{1- \frac{1}{x}} = x \left( 1+ \frac{1}{x} + \frac{1}{x^2} +\frac{1}{x^3} + \cdots \right) = x + 1 + \frac{1}{x} + \frac{1}{x^2} + \cdots $$

However, I tried

Series[1/(1 - 1/x), {1/x, 0, 2}]

it doesn't work

or

1/(1 - 1/x) /. 1/x -> t

Series[%, {t, 0, 3}]

% /. t -> 1/x

Expand[x*%]

it ends up to

$x \left( 1+ \frac{1}{x} + \frac{1}{x^2} +\frac{1}{x^3} + \cdots \right) $ cannot be expanded to break the bracket.

How should I do to arrive at $x + 1 + \frac{1}{x} + \frac{1}{x^2} + \cdots $?

You can add //Normal after Series[...]; however is this an expansion for small or for large x ? — b.gates.you.know.what, Aug 02 '14 at 10:06
Thank you very much! It works now. The expansion is for large $x$, such that $|1/x| <<1$ — Guest25874, Aug 02 '14 at 10:17

score 5 · Accepted Answer · answered Aug 02 '14 at 10:19

You can expand around infinity as follows:

 Series[x/(1 - 1/x), {x, Infinity, 2}]

$x + 1 + \frac{1}{x} + \left(\frac{1}{x}\right)^2 + \mathcal{O}\left[\left(\frac{1}{x}\right)^3\right]$

This returns a SeriesData expression, so you might want to add in a call to Normal as b.gatessucks suggests.

Series expansion for $\frac{x}{1- \frac{1}{x}}$

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