Depends if you want to input the series and get the sum of the series (in that case follow Eric's answer), or the converse, in which case you can ask for the series expansion.
sage: f = 1/(1-x)
sage: f.series(x)
1 + 1*x + 1*x^2 + 1*x^3 + 1*x^4 + 1*x^5 + 1*x^6 + 1*x^7 + 1*x^8
+ 1*x^9 + 1*x^10 + 1*x^11 + 1*x^12 + 1*x^13 + 1*x^14 + 1*x^15
+ 1*x^16 + 1*x^17 + 1*x^18 + 1*x^19 + Order(x^20)
Like this:
sage: assume(abs(x)<1)
sage: n = var('n')
sage: sum(x^n, n, 0, oo)
-1/(x - 1)
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.
Asked: 2018-08-11 17:14:11 +0100
Seen: 576 times
Last updated: Aug 13 '18
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Note that the question is about 1/(1 - x), not (1/1) - x.