Closed form of $k$-th term in the recurrence $D_j = 1 + z^2\left(1-\frac{1}{D_{j-1}}\right)$, where $D_0 = 1+z^2$

Question

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?

Answer 1

To expand upon my comment to @BarryCipra's answer ...

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Multiplying numerators and denominators by $(1-z^2)$ gives

$$D_0 = \frac{1 - z^4}{1 - z^2} \qquad D_1 = \frac{1 - z^6}{1-z^4} \qquad D_2 = \frac{1-z^8}{1-z^6}$$

We can show that the pattern continues.

Assuming

$$D_j = \frac{1 - z^{2j+4}}{1-z^{2j+2}} \tag{$\star$}$$

we have $$ \begin{align} D_{j+1} &= 1 + z^2\left(\; 1 - \frac{1 - z^{2j+2}}{1 - z^{2j+4}} \;\right) \\[6pt] &= 1 + z^2 \; \frac{z^{2j+2} - z^{2j+4}}{1-z^{2j+4}} \\[6pt] &=\frac{1 - z^{2j+4} + z^{2j+4} -z^{2j+6}}{1-z^{2j+4}}\\[6pt] &=\frac{1 -z^{2(j+1)+4}}{1-z^{2(j+1)+2}} \end{align}$$ so that $(\star)$ holds for all $j$, by induction. $\square$

Answer 2

If you work out the first few $D_j$'s explicitly, you should see a pattern:

$$D_1=1+z^2\left(1-{1\over1+z^2}\right)=1+z^2\left(z^2\over1+z^2\right)={1+z^2+z^4\over1+z^2}$$

$$D_2=1+z^2\left(1-{1+z^2\over1+z^2+z^4}\right)=1+z^2\left(z^4\over1+z^2+z^4\right)={1+z^2+z^4+z^6\over1+z^2+z^4}$$