What is $1+x+x^2+x^3...$? What is the difference between these two series? 
$$
\begin{align}
1+x+x^2+x^3+...+x^n+\mathcal O(x^{n+1})&=\frac{1}{1-x}\\
\\
1+r+r^2+r^3+...+r^{n-1}&=\frac{r^n-1}{r-1}\\
\end{align}
$$
I can't wrap my head around it; they both start with $1+x+x^2+x^3...$? Of course the first is the maclaurin series. But other than that, why don't they both yield the same result?
Thanks!
 A: Hint:
$$1+x+x^2+\dots+x^n=\frac{x^{n+1}-1}{x-1}$$
$$\frac{x^{n+1}-1}{x-1}=\frac{1-x^{n+1}}{1-x}=\frac1{1-x}-\frac{x^{n+1}}{1-x}=\frac1{1-x}+\mathcal O(x^{n+1})$$
which holds as $x\to0$, since $1-x\to1$.
Move $\mathcal O(x^{n+1})$ to the other side and you'll get
$$1+x+x^2+\dots+x^n+\mathcal O(x^{n+1})=\frac1{1-x}$$
A: Beware that "$=$" is not a symmetric relation in the presence of asymptotic notation. So while it is true that
$$\frac{1}{1-x} = 1+x+x^2+x^3+...+x^n+O(x^{n+1})$$
as $x\to 0$, we do not usually say that
$$1+x+x^2+x^3+...+x^n+O(x^{n+1})=\frac{1}{1-x}$$
because the left-hand side of that can be a lot of things other than $\frac{1}{1-x}$.
Some consider the conventional use of $=$ in these contexts to be misguided and misleading, and would rather we write
$$\frac{1}{1-x} \in 1+x+x^2+x^3+...+x^n+\mathcal O(x^{n+1})$$
whose right-hand side is interpreted as a set of functions.
A: The second (finite) series is a particularization of the first (with a shifted index):
$$
1 + x + x^{2} + x^{3} + \cdots + x^{n} + O(x^{n+1})= \frac{1}{1 - x}
$$
means: There exists constant $C > 0$ and $\delta > 0$ such that if $|x| < \delta$, then
$$
\left|\frac{1}{1 - x} - (1 + x + x^{2} + x^{3} + \cdots + x^{n})\right| \leq C|x|^{n+1}.
\tag{1}
$$
By contrast, the second (with $n$ replaced by $n + 1$) reads
$$
1 + r + r^{2} + r^{3} + \cdots + r^{n}
= \frac{r^{n+1} - 1}{r - 1}
= \frac{1 - r^{n+1}}{1 - r}
= \frac{1}{1 - r} - \frac{r^{n+1}}{1 - r}.
\tag{2}
$$
Equation (2) gives the specific form of the $O(x^{n+1})$, namely $\dfrac{x^{n+1}}{1 - x}$.
A: The two series actually do produce the same result. You just have to compare them fairly.
To begin with, you've listed $n$ terms in the first series 
(not including the "big-O" notation) and only $n-1$ terms in the second series. Let's make both series have $n$ terms, as follows.
(I'm also putting the "closed expression" side of the equation on the left
in order to have the big-O on the right, as is the standard practice.) 
\begin{align}
\frac{1}{1-x} &= 1+x+x^2+x^3+...+x^n+\mathcal O\left(x^{n+1}\right),\\
\frac{r^{n+1}-1}{r-1} &= 1+r+r^2+r^3+...+r^n.
\end{align}
Notice that $\frac{(-y)}{(-y)} = \frac yy,$ so if we reverse the subtraction
on the top and bottom of $\frac{r^{n+1}-1}{r-1}$ at the same time,
we get the same result:
$$
\frac{1 - r^{n+1}}{1 - r} = 1+r+r^2+r^3+...+r^n.
$$
Now distribute the numerator of $\frac{1 - r^{n+1}}{1 - r}$ over the
denominator:
$$
\frac{1}{1 - r} - \frac{r^{n+1}}{1 - r} = 1+r+r^2+r^3+...+r^n.
$$
Add $\frac{r^{n+1}}{1 - r}$ to both sides:
$$
\frac{1}{1 - r} = 1+r+r^2+r^3+...+r^n + \frac{r^{n+1}}{1 - r}.
$$
The first equation occurs in a context where $x$ is very small
(close to zero), so that the $\mathcal O\left(x^{n+1}\right)$ term
represents just an error correction to the previous terms.
If we likewise suppose that $\lvert r\rvert \ll 1,$ so that we are
comparing apples to apples (both series dealing with the same kind of numbers), then 
$\frac{r^{n+1}}{1 - r} = \mathcal O\left(r^{n+1}\right),$
and we can write
$$
\frac{1}{1 - r} = 1+r+r^2+r^3+...+r^n + \mathcal O\left(r^{n+1}\right).
$$
So this is the same as the first equation, but with the variable $r$
substituted for $x.$
