Can a sequence contain an undefined term (e.g. a member acquired by dividing a number by zero)? I learned about the theorem that every convergent series is bounded.
However, take $a_n = \frac{1}{n-3}$ with $n=3$. 
It is unbounded, since $\sup{a_n} = +\infty$ (EDIT: it is not.). 
And yet, it is convergent with $\lim {a_n} = 0$
Am I violating some axiom of sequences with this one?

A convergent sequence of real numbers is bounded. It's important to recognize that $\infty$ and $-\infty$ are not real numbers.
There are extensions of the real numbers in which these symbols become legitimate numbers, although even in those systems there are some restrictions on the operations that can be performed using $\infty$ and $-\infty$.
For example, the extended real numbers, often denoted $[-\infty, \infty]$ includes all of $\mathbb R$ along with $\infty$ and $-\infty$, with the expected order ($-\infty < x < \infty$ for any real number $x$), and some operations involving infinity are allowed, such as $\infty + x = \infty$ for any real number $x$. However, expressions such as $\infty - \infty$ are still undefined. Similarly, $1/0$ is undefined in this system, largely because $\lim_{x \to 0}1/x$ doesn't exist: the one-sided limit from the left is $-\infty$ and the one-sided limit from the right is $\infty$, just as in the real numbers.
Another system is the extended nonnegative real numbers, often denoted $[0,\infty]$. This is very useful in measure theory, for example, where we are measuring the sizes of sets and want to include $\infty$ as an allowable size. In this system, $1/0$ is often defined to be $\infty$, because there's no ambiguity about $\lim_{x \to 0}1/x$: we can only approach from the right since only nonnegative values are available.
As a side note, the notion of convergence in $[-\infty,\infty]$ and $[0,\infty]$ is also extended to include $\infty$ as a valid limit. For example, the sequence $a_n = n$ converges to $\infty$ in both cases, yielding another example of a convergent unbounded sequence. Similarly, in $[-\infty, \infty]$, the sequence $a_n = -n$ converges to $-\infty$.
Returning to your example: in either $[0,\infty]$ or $[-\infty, \infty]$, there's no problem explicitly defining the sequence $(a_n)_{n=1}^{\infty} = -1/2, -1, \infty, 1, 1/2, 1/3, 1/4, \ldots$ and showing that it converges to $0$. And it's clearly unbounded since one of the terms is $\infty$. The only issue with your example is your use of a formula to generate the $\infty$ at $a_3$, because division by zero may or may not make sense depending on what number system you're using.
