I did the inverse z-transform of $\frac{2}{z+1/3}$ and got a result, by doing so
$$\mathscr{Z}^{-1}\{\frac{2}{z+1/3}\}=2\lim_{z_0\longrightarrow -1/3}(z+1/3)\frac{z^{n-1}}{z+1/3}=2\lim_{z_0\longrightarrow -1/3} z^{n-1}=2(-1/3)^{n-1}$$
But another source, claims this is the right result:
$$2(-1/3)^{n-1}\Theta(n-1)$$
Why should this result with the Heaviside function be correct, when the formula for the inverse transform is given by the contour integral:
$$\frac{1}{2\pi i}\oint_C x(z)z^{n-1}dz$$
Thanks