<s>[INST] 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. 
 [/INST] 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.
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