Something I'm not getting about geometric series. The notation for geometric series is
$$\sum_{n=1}^{\infty}{a_0r^{n-1}} ={a_0}+{a_0}r+{a_0}r^2+{a_0}r^3+...  $$
and a geometric series will converge if $ -1<r<1 $ or if $ {a_0}=0 $.
In math class some time ago the teacher gave us the following problem:

With what values of $x$ will the following series converge?
  $$ \sum_{n=1}^{\infty}(\frac{3x}{x-4})^{n-1} $$

I tried to solve this by checking when $ -1<r<1 $ and when $ {a_0}=0 $, because those are the cases when it'll converge, but the teacher told me that $ {a_0}\neq0 $ because $0^0$ is undefined.
There is something I'm missing. What even actually is the ${a_0}$ in this case? And what is the $r$?
 A: Here $r=\frac{3x}{x-4}$ and $a_0=1$. Here and in a number of other places it is convenient to define $0^0=1$, so that everything makes sense when $r$ is anything in $(-1,1)$. The only reason I know of to justify not always saying $0^0=1$ is that $f(x,y)=x^y$ isn't continuous at $(0,0)$. Two examples to see this are $\lim_{x \to 0^+} x^x = 1$ but $\lim_{x \to 0^+} x^{-\frac{1}{\ln(x)}}=1/e$.
A: In this case, that is, in your problem,
$a_0=1$ and $r=\frac{3x}{x-4}$
Now the series will be convergent iff $$|r|<1$$
$$-1 \le r \le 1$$
$$-1 \le \frac{3x}{x-4} \le 1$$
We divide this into $2$ parts:

$4-x \le 3x \le x-4\,$ when $x-4>0$
and 
$\, 4-x\ge 3x \ge x-4 $ when $x-4<0$

From the first part, we get that 

$x \ge 1$ and $x \le -2$ and $x>4$.


From the second part, we get that 

$x \le 1$ and $x \ge -2$ and $x<4$.

But there is no such $x \in \mathbb{R}$ which satisfies all $3$ inequalities in the first part.
However, for $x \in [-2,1]$, all $3$ inequalities in the second part are satisfied and the value of $r$ lies between $-1$ and $1$.
Hence the series does converge for $x \in [-2,1]$.
