Recall that for [tex]|z|<1[/tex], we have
[tex]\dfrac1{1-z}=\displaystyle\sum_{n\ge0}z^n[/tex]
Then
[tex]\dfrac1{1-z}=-\dfrac1z\dfrac1{1-\frac1z}=-\dfrac1z\displaystyle\sum_{n\ge0}z^{-n}=-\sum_{n\ge0}z^{-n-1}[/tex]
valid for [tex]|z|>1[/tex], so that
[tex]\dfrac1{1-z^2}=-\dfrac1{z^2}\dfrac1{1-\frac1{z^2}}=-\dfrac1{z^2}\displaystyle\sum_{n\ge0}z^{-2n}=-\sum_{n\ge0}z^{-2n-2}[/tex]
also valid for [tex]|z|>1[/tex].
