Closed form of $k$-th term in the recurrence $D_j = 1 + z^2\left(1-\frac{1}{D_{j-1}}\right)$, where $D_0 = 1+z^2$ I have $D_0=1+z^2$ and $D_{j}=1+z^2(1-1/D_{j-1})$ for $j>0$. I want to write an expression for arbitrary $D_k$, and what I have is below.
$D_{k}=1+z^2\Bigg(1-\bigg(1+z^2\big(1-(1+z^2(1-(\cdots(1+z^2)^{-1}\cdots)^{-1})^{-1})^{-1}\big)^{-1}\bigg)^{-1}\Bigg)$
Is there better notation I can use to make this more precise/clear?

If you work out the first few $D_j$'s explicitly, you should see a pattern:
$$D_1=1+z^2\left(1-{1\over1+z^2}\right)=1+z^2\left(z^2\over1+z^2\right)={1+z^2+z^4\over1+z^2}$$
$$D_2=1+z^2\left(1-{1+z^2\over1+z^2+z^4}\right)=1+z^2\left(z^4\over1+z^2+z^4\right)={1+z^2+z^4+z^6\over1+z^2+z^4}$$
