BIBO stability and input at system's pole My understanding of BIBO stability is that if a bounded input is applied to a discrete-time, linear, time-invariant BIBO system, the output will also be bounded. A sufficient condition for this is that the poles of the transfer function are within the unit circle.

Suppose I have a system with the transfer function 
$$
H(z)={(1−2z^{−1})(1−{\frac{1}{6}} z^{−1}) \over (1−\frac{1}{2}z^{−1})(1−\frac{1}{3}z^{−1})}
$$
The poles of this system are within the unit circle and so is a BIBO stable system.

If I apply an input of $x[n]=(\frac{1}{3})^n$, which is bounded, the output would involve $H(\frac{1}{3})=\infty$, which is unbounded - i.e. a bounded input produced an unbounded output.

Obviously, there is something amiss in my understanding here, but what?
The signal $x[n] = (\frac{1}{3})^n$ is not bounded. As $n \to -\infty$, $x[n] \to \infty$; therefore, the system's response to your candidate signal $x[n]$ has no bearing on its BIBO stability.

A bounded, causal version of the signal (and likely what you were thinking) is $x_b[n] = (\frac{1}{3})^n u[n]$, where $u[n]$ is the unit step function. The response of the system to this signal is bounded as you would expect.