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!

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 $x.$
