`slelièvre`

's answer gives a *finite* series, completed byy an O(n) term.. If you want the (supposedly exact) *infinite* series, you have (at least) two possibilities :

Maxima's `powerseries`

does the job :

sage: (1/(1-x)).maxima_methods().powerseries(x,0)

sum(x^i1, i1, 0, +Infinity)

Sympy''s `summation`

does it also :

sage: sympy.summation(x^j,(j,0,oo))

Piecewise((1/(-x + 1), Abs(x) < 1), (Sum(x**j, (j, 0, oo)), True))

Sympy's solution is better (it gives the conditions of validity of the result), but currently can't be automatically transtaled to Sage, because Sympy's `Sum`

method does not (yet) have a `_sage_()`

method.

Note that the question is about 1/(1 - x), not (1/1) - x.