Z-Transform of $x(n) = 3^n$ First of all, thank you all for your answers. I know the z transform for

$$ x(n)=3^n \space ; \space  n\geqslant  3 $$ 

or rather 

$$ x(n)= 3^n u(n-3) $$$$\begin{align}X(z)&amp;=\sum_{n=-\infty}^{\infty}x[n]z^{-n}\\&amp;=\sum_{n=-\infty}^\infty3^nu(n-3)z^{-n}\\&amp; \text{change of variables: } n'=n-3 \\&amp; =\sum_{n=-\infty}^\infty 3^{(n'+3)} u(n') z^{(-n'-3)}\\&amp; \text{recall:  } n'-&gt; n \\&amp;=3^3z^{-3}\sum_{n=0}^{\infty}(3 z^{-1})^{n}\\&amp; X(z) =3^3 z^{-3}\frac{1}{1-3{z^{-1}}},\quad |z|&gt;3  \tag{1}\end{align} $$\begin{align} \\&amp;  \text{instead, for the sequence: } x(n)=3^{n} \ ; n\leqslant -3  \\&amp;=\sum_{n=-\infty}^\infty3^nu(-n-3)z^{-n}\\&amp; \text{change of variables: } n'=-n-3 \\&amp; =\sum_{n=-\infty}^\infty 3^{(-n'-3)} u(n') z^{-(-n'-3)}\\&amp; \text{recall:  } n'-&gt; n \\&amp;=3^{-3} z^{3}\sum_{n=0}^{\infty}(3^{-1} z)^{n}\\&amp;  X(Z)= 3^{-3}z^{3}\frac{1}{1- \frac{z}{3}} \quad |z|&lt;3\tag{2}\end{align}
Is it correct?
And for @Matt L. I don't understand your mathematical step initial when you write: 
$$\begin {align} \sum_{n=-\infty}^3 3^nz^{-n}  =\sum_{n=-\infty}^0 3^{(n+3)} u(n') z^{-(n+3)} \end {align} $$ 

Can you write me the generic formula?
The signal $x[n]=3^n$ for $n\le -3$ (and zero otherwise, I assume) can be written as

$$x[n]=3^nu[-n-3]\tag{1}$$

To compute its $\mathcal{Z}$-transform you simply need to use the formula:

$$\begin{align}X(z)&amp;=\sum_{n=-\infty}^{\infty}x[n]z^{-n}\\&amp;=\sum_{n=-\infty}^{-3}3^nz^{-n}\\&amp;=\sum_{n=-\infty}^03^{(n-3)}z^{-(n-3)}\\&amp;=3^{-3}z^{3}\sum_{n=0}^{\infty}\left(\frac{z}{3}\right)^n\\&amp;=3^{-3}z^{3}\frac{1}{1-\frac13 z},\quad |z|&lt;3\tag{2}\end{align}$$

It's important to specify the region of convergence ($|z|&lt;3$), otherwise the $\mathcal{Z}$-transform does not uniquely identify the corresponding sequence $x[n]$.