<s>[INST] 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!
 [/INST] 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 $x.$
</s>