How does $ \sum \limits_{n=0}^\infty\left (\frac{1}{z^2}\right)^m = \frac{z^2}{z^2-1}$?

Question

This was given by an exercise's answer, but the steps in between was not given, so I was wondering how $$\sum \limits_{n=0}^\infty \left(\frac{1}{z^2}\right)^n=\frac{z^2}{z^2-1}$$ The exercise was related to $z$-transform.

Answer 1

Just simplify: \begin{equation} \frac{z^2}{z^2-1}=\frac{1}{1-\frac{1}{z^2}}. \end{equation} Using the geometric series formula $$ \sum_{n=0}^\infty ar^n=\frac{a}{1-r}$$ for $\lvert r\rvert<1$, we find that taking $r=\frac{1}{z^2}$, for $\lvert\frac{1}{z^2}\rvert<1$ we have \begin{equation} \frac{z^2}{z^2-1}=\frac{1}{1-\frac{1}{z^2}}=\sum_{n=0}^\infty \frac{1}{z^{2n}}. \end{equation}

Answer 2

You may use the infinite geometric series formula: $$\sum_{k=0}^\infty ar^k=\lim_{n\to \infty}\sum_{k=0}^n ar^k=\lim_{n\to \infty}\frac{a(1-r^{n+1})}{1-r}=\frac{a}{1-r}-\lim_{n\to \infty}\frac{a(r^{n+1})}{1-r}=\frac{a}{1-r}$$ $\because$ $r^{n+1} \to 0$ when $n\to \infty$ since $ \vert r\vert \lt1$

In your case, $a=1$ and $\frac {1}{z^2}$. Hence, assuming $\vert z\vert \gt 1$, $$\sum_{k=0}^\infty \left(\frac{1}{z^2}\right)^k=\frac{1}{1-\frac{1}{r^2}}=\frac{r^2}{r^2-1}$$