Find upper and lower bound when $a$ is negative Let $\alpha \neq 1$. I showed that $1+\alpha+ \dots+\alpha^n=\frac{1-\alpha^{n+1}}{1-\alpha}$.
I want to show that when $0<|\alpha|<1$, that the set $\{1+\alpha+\dots+\alpha^n \mid n \in \mathbb{N} \}$ is bounded. And I want to find the least upper bound.
When $0<\alpha<1$, we have that $\alpha^{n+1}<1 \Rightarrow \frac{1-\alpha^{n+1}}{1-\alpha}>0$.
Also $\alpha^{n+1}>0 \Rightarrow \frac{1-\alpha^{n+1}}{1-\alpha}<\frac{1}{1-\alpha}$.
In order to show that $\frac{1}{1-\alpha}$ is the least upper bound we suppose that there is some $\epsilon>0$ such that $\frac{1}{1-\alpha}-\epsilon$ is a lower bound and we find a contradiction.
Can we find in the same manner the lower and upper bound of the set when $-1<0<\alpha$ ?
Let $-1<a<0 $. We set $\beta=-\alpha$. Then $0<\beta<1$.
Then we have the following.
$1+\dots+\alpha^n=\left\{\begin{matrix}
\frac{1+\beta^{n+1}}{1+\beta} &, \text{ if n is even} \\ 
\frac{1-\beta^{n+1}}{1+\beta} &, \text{ if n is odd} 
\end{matrix}\right.$
We have that $\beta>0 \Rightarrow \beta^{n+1}>0 \Rightarrow \frac{1+\beta^{n+1}}{1+\beta}>\frac{1}{1+\beta}$.
Also $\beta<1 \Rightarrow \beta^{n+1}<\beta \Rightarrow \frac{1+\beta^{n+1}}{1+\beta}<1$.
Furthermore, we have that $\frac{1-\beta^{n+1}}{1+\beta}<\frac{1}{1+\beta}$ and $\frac{1-\beta^{n+1}}{1+\beta}>0$.
Do we use these upper and lower bounds in order to find the lower and least upper bound of the set when $-1<\alpha<0$ ?
 A: It often helps to look at a specific example, where you can do some calculations and pin things down.  In this case, we're looking generally at sets of the form $\{1+\alpha+\alpha^2+\cdots+\alpha^n\mid n\in\mathbb{N}\}$ with $-1\lt\alpha\lt1$. So let's pick a couple of nice, easy values of $\alpha$ to look at.
Even though you did the positive $\alpha$ case correctly, it won't hurt to look at $\alpha={1\over2}$, for which the set is
$$\left\{1+{1\over2}+\left(1\over2\right)^2+\cdots+\left(1\over2\right)^n\mid n\in\mathbb{N}\right\}=\left\{1,{3\over2},{7\over4},{15\over8},{31\over16},\ldots \right\}$$
It's easy to see here that the lower bound is simply $1$, and the least upper bound is $2$, which, of course, is equal to $1\over1-\alpha$ with $\alpha={1\over2}$.
Now look at $\alpha=-{1\over2}$.  The corresponding set now is
$$\left\{1,{1\over2},{3\over4},{5\over8},{11\over16},\ldots \right\}$$
For this set, the least upper bound is $1$ and the greatest lower bound is ${1\over2}$.  That's because after the first two entries in the list, each new entry is between the previous two. A bit of thought should convince you that that holds whenever you are alternately adding and subtracting numbers that are getting smaller and smaller, which is the case with $\alpha^n$ when $-1\lt\alpha\lt0$. So in general, if $-1\lt\alpha\lt0$, the upper and lower bounds are simply $1$ and $1+\alpha$.
