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. 
 A: The expression
$$ 1 + x + x^2 + x^3 + \dotsb = \sum_{k=0}^{\infty} x^k $$
is a geometric series.  We say that a series converges if the corresponding sequence of partial sums converges.  That is
$$ \sum_{k=0}^{\infty} x^k \text{ converges}\iff \lim_{n\to\infty} \sum_{k=0}^{n} x^k =: S_n < \infty.$$
But note that
$$ S_n = 1 + x + x^2 + x^3 + \dotsb + x^n
\implies x S_n = x + x^2 + x^3 + x^4 + \dotsb + x^{n+1}.$$
Subtracting and canceling terms as appropriate, this gives us
$$ (1-x)S_n = S_n - xS_n = 1 - x^{n+1}
\implies S_n = \frac{1-x^{n+1}}{1-x}.$$
Taking the limit, we get
$$ \lim_{n\to\infty} S_n
= \begin{cases}
\frac{1}{1-x} & \text{ if $|x|<1$ (since $x^{n+1} \to 0$),} \\
\infty & \text{ if $|x|>1$ (since $|x|^{n+1} \to \infty$), and} \\
\text{indeterminate} & \text{otherwise.}
\end{cases}
$$
Now, observe that if $|x|<1$, then
$$ f(x) = 1 + \sum_{k=2}^{\infty} x^k = \left(\sum_{k=0}^{\infty} x^k\right) - x = \frac{1}{1-x} - x.$$
For other values of $x$, the series won't converge, and the function will be undefined.  Hence the domain of $f$ (presuming that the domain is real, and not complex) is the interval $(-1,1)$, and on this domain it is given by
$$ f(x) = \frac{1}{1-x} - x = \frac{1 - x(1-x)}{1-x} = \frac{1 - x + x^2}{1-x} = 1 + \frac{x^2}{1-x}.$$
Finally, suppose that $y$ is in the range of $f$.  Then there is some $x\in (-1,1)$ such that $y=f(x)$.  Solving, we get
$$ y = \frac{ 1-x + x^2}{1-x}
\implies (1-x)y = y - yx = 1-x+x^2
\implies 0 = x^2 + (y-1)x + (1-y).
$$
Thus
$$ x = \frac{1 - y \pm \sqrt{ (y-1)^2 - 4(1-y)}}{2}, $$
and so the equation $y = f(x)$ has a real solution $x$ as long as
$$ (y-1)^2 - 4(1-y)
= y^2 - 2y + 1 - 4 + 4y
= y^2 + 2y - 3
= (y+3)(y-1)
\ge 0. $$
This happens whenever $y\le-3$ or $y\ge 1$, i.e.
$$ y \in (-\infty,-3]\cup [1,\infty). $$
However, note that we also require $-1 < x < 1$.  For such $x$, we have $1-x > 0$, and so
$$ y = 1 + \frac{x^2}{1-x} \ge 0.$$
Therefore the range of $f$ is contained in the set
$$ \Big[ (-\infty,-3]\cup [1,\infty) \Big] \cap [0,\infty)
= [1,\infty).$$
Slightly more carefully, note that $f$ is continuous on $(-1,1)$, that $f(0) = 0$, and that
$$\lim_{x\to 1^{-}} f(x) = \infty.$$
This is sufficient to show that $[1,\infty)$ is contained in the range of $f$, which finally allows us to conclude that the range of $f$ is
$[1,\infty).$

Something worth noting is that the domain of $f$ is the set $(-1,1)$, since $f$ is only defined when the series is convergent.  However, as we observed above, on this domain, we have the identity
$$
f(x) = 1 + \sum_{k=2}^{\infty} x^k = 1 + \frac{x^2}{1-x}.
$$
The right-most expression defines a legitimate function on $\mathbb{R} \setminus \{1\}$.  However, this is not the same function as $f$, since $f$ was defined by the series, and not by the rational expression on the right.  We can, however, think of the function
$$ g(x) = 1 + \frac{x^2}{1-x} $$
as an extension of $f$ to a larger domain.  Indeed, in the context of complex analysis (a branch of mathematics that deals with functions that have domains in the complex plane), it can be shown that this extension is (in some sense) the "right" extension.
A: 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.
A: 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 $[1,\infty)$.
