ROC of Z Transform of $x(n) = 2(3)^nu(-n)$ Using definition, I got its Z transform as $X(z) = \dfrac{2}{1-\dfrac{z}{3}}$ and the summation converges only when $|z|&lt;\frac{1}{3}$. So its ROC is $|z|&lt;\frac{1}{3}$.

But my question is: for such a left sided signal $x(n)$, its ROC should be inner to the innermost pole. But the pole of $\dfrac{2}{1-\dfrac{z}{3}}$ is $z=3$ and not $z= \frac{1}{3}$. So the ROC I found is incorrect, right?
Using the definition, and letting $k=-n$:

$X(z)=2 \sum_{n=-\infty}^03^nz^{-n}=2\sum_{k=0}^{\infty}\big(\frac{1}{3}z\big)^k$

Now we can use the geometric series (https://en.wikipedia.org/wiki/Geometric_series) formula only if we require $|\frac{1}{3}z|&lt;1$, or equivalently $|z|&lt;3$