Let $z_n$ be a sequence of numbers such that $$z_{n+1} \leq cz_n^{1+\epsilon}$$ for a constant $c>0$ and $\epsilon > 0$.
If $z_0$ is sufficiently small, is it possible to say that $$z_n \leq f(n)$$ where $f(n) \to 0$ as $n \to \infty$? $f$ may depend on $\epsilon$ and $z_0$.
Iterating the inequality we get $$\begin{align*} z_{n+1}&\le c\,z_n^{1+\epsilon}\\ &\le c^{1+(1+\epsilon)}\,z_{n-1}^{(1+\epsilon)^2}\\ &\le c^{1+(1+\epsilon)+(1+\epsilon)^2}\,z_{n-2}^{(1+\epsilon)^3}\\ &\le\dots\\ &\le c^{1+(1+\epsilon)+(1+\epsilon)^2+\dots+(1+\epsilon)^n}\,z_0^{(1+\epsilon)^{n+1}}\\ &=c^{\tfrac{(1+\epsilon)^n-1}{\epsilon}}\,z_0^{(1+\epsilon)^{n+1}}\\ &=c^{-1/\epsilon}\,z_0^{1+\epsilon}\bigl(c^{1/\epsilon}\,z_0\bigr)^{(1+\epsilon)^n}. \end{align*}$$ If $c^{1/\epsilon}\,z_0<1$, then $z_n\to0$.