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. 

Let $f(x)$ be the given function.

Ignoring the first term, the series is a geometric series with ratio $x$, hence converges if and only if $|x| < 1$. It follows that the domain of $f$ is the interval $(-1,1)$.

For $x \in (-1,1)$, the geometric series sums to $\large{\frac{x^2}{1-x}}$, hence
$$f(x) = 1 +\frac{x^2}{1-x},\;\;\;\text{for}\,-1 < x < 1$$
Taking the derivative and simplifying the result, we get
\begin{align*}
f'(x) &= \frac{x(2-x)}{(1-x)^2}\\[8pt]
&\text{hence}\\[8pt]
f'(x) &<0\;\;\;\text{if$\;\;-1 < x <0$}\\[4pt]
f'(0) &=0\\[4pt]
f'(x) &>0\;\;\;\text{if$\;\;0<x<1$}\\[4pt] 
\end{align*}
It follows that $f$ is decreasing on the interval $(-1,0]$, and increasing on the interval $[0,1)$.

Thus, the minimum value of $f$ is $f(0)=1$.

Noting that $f(x)$ can be expressed as
$$f(x) = -x + \frac{1}{1-x}$$
we get
$$\lim_{x=1^{-}}f(x) = -1+\infty = \infty$$
Hence, since $f$ is continuous on its domain, it follows that the range of $f$ is $[1,\infty)$.
