$A_k\leq C [2^{2k}A_{k-1}]^{\mu+1}, \ k=0,1,...$ implies $A_k\to 0$?

Question

Consider the nonlinear recursive relation $$A_k\leq C [2^{2k}A_{k-1}]^{\mu+1}, \ k=0,1,...$$

where $C,A_k,\mu>0$. How can one show that if $A_0$ is small, then $A_k\to 0$?

Thanks.

Answer 1

Write the recursion on the form $$\frac{A_k}{A_{k-1}}\le C(\alpha^kA_{k-1})^\mu$$ for $\alpha=4^{(\mu+1)/\mu}$. This can be rewritten as $$\frac{\alpha^kA_k}{\alpha^{k-1}A_{k-1}}\le D(\alpha^{k-1}A_{k-1})^\mu$$ with $D=C\alpha^{1+\mu}$. Now ensure that $DA_0^\mu<1$, and show by induction that $(\alpha^kA_k)$ is a decreasing sequence.