domain and range of function $y= 1+\sum_{n=2}^\infty x^n$ My friend give me a question to find domain and range  of
$y= 1+\sum_{n=2}^\infty x^n$
there is no more description about the problem, so  I think the domain of that function is all  $\mathbb{R} $ and range of that function is $ \mathbb{R}$ too
but he told me that I was wrong, domain of that function is $-1 < x < 1 $ and range of that function is $y \leq -3  $ or $y  \geq 1  $  that was shocked me.
I never knew whether domain of convergence or divergence function form will be an answer of this function.
Moreover, according to  WFA  they said $y= 1+\sum_{n=2}^\infty x^n$ 
 equal to $y =  1 - \frac{x^2}{x-1}$ for  $\left | x \right | <1$  so domain of function should be   $ \left \{ x\in \mathbb{R} : x\neq 1 \right \}$, isn't it?
Which one got the right answer and what is real definition of domain and range
Sorry for my english, Thank you in advance. 

If all you've got is $\text{“} 1 + x^2+x^3+x^4+\cdots\text{''},$ then the question of what the domain is is problematic in several respects.
The simplest of those may be whether $x$ is supposed to be a real number or a complex number. And perhaps it could also be a matrix or any of a variety of other things. Next there is the question of whether the value of the function is just supposed to be the ordinary sum of the series, with the usual kind of convergence. There are other sorts of convergence from the one you first learn.
However, if we just assume $x$ is supposed to be a real number and that the value of the function is the sum of the series with the most usual sort of convergence, then we confront the fact that the series converges only if $-1<x<1.$ For example, if $x=1$ then the series becomes $1+1+1+\cdots=\infty.$ Should we say it's undefined when the sum is $\infty$ or should we just say it's defined and its value is $\infty\text{?}$ Here again, if we assume values have to be real numbers, we have to exclude this.
The sum of a geometric series is given by
$$
a + ar + ar^2 + ar^3 + ar^4 + \cdots = \frac a {1-r}, \text{ if } -1<r<1.
$$
In this case we have $a=x^2$ and $r=x,$ so
$$
x^2 + x^3 + x^4 + \cdots = \frac{x^2}{1-x} \text{ if } -1<x<1.
$$
As for the range, you missed something if you just assumed it was $\mathbb R.$ Generally it is more work to find the range than the domain.
Notice that as $x$ approaches $1$ from below, $x^2$ approaches $1$ and $1-x$ approaches $0$ from above, so $x^2/(1-x)$ approaches $+\infty.$
Notice that $1-x$ is positive when $-1<x<1,$ so $x^2/(1-x)$ is positive except when $x=0.$ So the whole of $[0,\infty)$ is in the range of $x\mapsto x^2/(1-x).$ So you get $[1,\infty)$ as the range of the function you started with.
