Solve Inverse Z-Transform for H(z)=1/(1-2/3*z-1)

[Drao92]

Homework Statement

H(z)=1/(1-2/3*z^-1)
h(n)=?

Homework Equations

I tried to use the residue method to solve this problem but it doesn't give me the good result and i am not sure if i don't know how to use the method or this problem can't be solved using this method.
Also, its pretty obvious that the result is h(n)=(2/3)ⁿ because it looks like a geometric progression

The Attempt at a Solution

I used the formula of residue method.
h(n)=residue(H(z)*z^n-1) when z=2/3
The result is 3*(2/3)ⁿ
Its this result wrong because when we calculate Z transform of (2/3)ⁿ which is (1-(2/3)ⁿ*z^-n )/(1-(2/3)*z^-1) we ignore (2/3)ⁿ because is 0 when n=infinity?

 

[vela]

Homework Statement

H(z)=1/(1-2/3*z^-1)
h(n)=?

Homework Equations

I tried to use the residue method to solve this problem but it doesn't give me the good result and i am not sure if i don't know how to use the method or this problem can't be solved using this method.
Also, its pretty obvious that the result is h(n)=(2/3)ⁿ because it looks like a geometric progression

The Attempt at a Solution

I used the formula of residue method.
h(n)=residue(H(z)*z^n-1) when z=2/3
The result is 3*(2/3)ⁿ
Its this result wrong because when we calculate Z transform of (2/3)ⁿ which is (1-(2/3)ⁿ*z^-n )/(1-(2/3)*z^-1) we ignore (2/3)ⁿ because is 0 when n=infinity?

You have
$$z^{n-1} H(z) = \frac{z^{n-1}}{1-\frac{2}{3z}} = \frac{z^n}{z-\frac{2}{3}}.$$ How did you get $h(n)=3(2/3)^n$ from that?

 

[Drao92]

I think i found what what I am doing wrong.
I made H(z)=3z/(3z-2) then i multiply by z^n-1 and simplify by (3z-2) but i must simplify by (3z-2)/3 when calculating the residue... From now ill stick only to z-x form for denominator.
Eveyrthing is clear now, thanks!