Complex series: $\sum_{n=0}^\infty\left( z^{n-2}/5^{n+1}\right)$ for $0 < |z| < 5$

How would one compute
$$\sum_{n=0}^\infty\frac{z^{n-2}}{5^{n+1}}$$
where $0\lt|z|\lt5$?

I have literally no idea where to start, all I know is that the answer will not have summations. Any help would be appreciated!

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As said above,

$$\sum_{n=0}^\infty {z^{n-2}\over 5^{n+1}}= {1\over 5z^2}\sum_{n=0}^\infty\left({z\over 5}\right)^n.$$

To solve this, it may aid you to make the substitution $u=\frac{z}{5}$. Then,
$${1\over 5z^2}\sum_{n=0}^\infty\left({z\over 5}\right)^n=\frac{1}{5z^2}\sum_{n=0}^{\infty}u^n.$$

The sum is geometric in $u$; thus, you apply the geometric series formula.

$$\sum_{n=0}^\infty {z^{n-2}\over 5^{n+1}}= {1\over 5z^2}\sum_{n=0}^\infty\left({z\over 5}\right)^n$$

Does this make it a bit more palatable?

HINT:

$$\frac{z^{n-2}}{5^{n+1}}=\frac1{5z^2}\left(\frac{z}5\right)^n$$