I deleted this at StackExchange as it seems advanced or not very relevant for there and I want to ask here.
Let z be a complex number, $Im(z)\neq 0$ or $Re(z)<4$, m be a natural number, let $f(z,m)$ be defined recursively;
$f(z, 0) = z$
$f(z, 1) = z^{z/2}$
$f(z,2)=\left(z^{z/2}\right)^{\frac{z^{z/2}}{2}}$
and so on,
$$f(z,n)=f(z,n-1)^{f(z,n-1)/2}$$
with
$$g(z,m)=\prod _{n=0}^m \left(1-f\left(\sqrt{z},n\right)\right)$$
then,
$$\sum _{m=0}^{\infty} f\left(\sqrt{z},m\right)g(z,m)=1-z-\sum _{m=0}^{\infty}(f\left(\sqrt{z},m+1\right)){}^{2}g(z,m)$$ with the first three terms of the sum on the left given by
$\left(1-\sqrt{z}\right) \sqrt{z}+\left(1-\sqrt{z}\right) z^{\frac{\sqrt{z}}{4}} \left(1-z^{\frac{\sqrt{z}}{4}}\right)+\left(1-\sqrt{z}\right) \left(z^{\frac{\sqrt{z}} {4}}\right)^{\frac{z^{\frac{\sqrt{z}}{4}}}{2}} \left(1-z^{\frac{\sqrt{z}}{4}}\right) \left(1-\left(z^{\frac{\sqrt{z}}{4}}\right)^{\frac{z^{\frac{\sqrt{z}}{4}}}{2}}\right) etc$
and the values at zero by taking the appropriate limit of each term.
What are the usual/standard ways to approach a problem of this kind? Is there a known method to establish it?
What is known about the convergence of such series?
Is there any immediate way to prove, or analytically establish the convergence of either identity?