<s>[INST] How to obtain a formula for $f(z)$ given this recurrence I am trying to figure out how to derive a formula for $f(z)$ that is a function of $z$ and maybe $k \in \mathbb{N}$:
$$
f(z) = 1+z f \bigg(\frac{z}{1+z} \bigg)
$$
As an attempt, I tried a change of variable $z=\frac{1}{x}$, and I get:
$$
f\bigg(\frac{1}{x}\bigg) = 1 + \frac{1}{x}f\bigg(\frac{1/x}{1+1/x}\bigg)
$$
which evaluates to the following after simplifying the terms inside $f$:
$$
f\bigg(\frac{1}{x}\bigg) = 1 + \frac{1}{x}f\bigg(\frac{1}{1+x}\bigg)
$$
Now, given the above, is the following telescoping operation valid?
$$
f\bigg(\frac{1}{x}\bigg) = 1 + \frac{1}{x}\bigg[ 1 + \frac{1}{x} + f\bigg(\frac{1}{2+x}\bigg) \bigg] = 1 + \frac{1}{x} + \frac{1}{(x+1)^2} + ... + \frac{1}{(x+k-3)^{k-2}} + \frac{1}{(x+k-2)^{k-1}}f\bigg(\frac{1}{k+x}\bigg)
$$
Also, how do I transform the above to a solution for $f(z)$?
 [/INST] In the last step you plug $1+x$ in the place of $x$, hence you should have 
$$ f\left( \frac{1}{x}\right) = 1 + \frac{1}{x}f\left( \frac{1}{1+x}\right) = $$
$$ 1 + \frac{1}{x}\left[ 1 + \frac{1}{1+x}f\left( \frac{1}{2+x} \right)\right]$$
So I think the telescopic series you achieved isn't correct.
</s>