Existence of initial conditions for the convergence of a recursively defined sequence to zero

Question

Motivation

Consider a sequence $m_k$ of non-negative real numbers satisfying the following recursion : $0<m_0 <1$ and there exists constants $C>0,D>1,q>1$ such that $$ m_{k+1} = C \times D^{k} \times m_{k}^q $$ This recursive constraint has a "non-autonomous" nature : $m_{k+1}$ is constrained not only by a function of $m_k$, but also by one involving $k$. This usually makes analysis difficult. Here, however, we can iterate the bound by doing $$ m_{k+1} = C D^km_k^q = CD^k(CD^{k-1}m_{k-1}^q)^q = C^{1+q}D^{k+(k-1)q}m_{k-1}^{q^2} $$ Now repeat this for $m_{k-1}$, and so on. Pulling this all the way and applying elementary estimates tells us that $$ m_{k+1} \leq (D^{\frac{q}{(q-1)^2}}C^{\frac 1{q-1}}m_0)^{q^k} C^{-\frac 1{q-1}} $$ In particular, if $\epsilon = D^{\frac{q}{(q-1)^2}}C^{\frac 1{q-1}}$, then $m_0 < \epsilon$ implies that $\lim_{k \to \infty} m_k = 0$. In particular, there is an $\epsilon>0$ such that $0<m_0<\epsilon$ implies $\lim_{k \to \infty}m_k = 0$. That property is what I desire for the next recurrence.

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Question

Now, $x^q$, for $q >1$, decays slightly faster than $\frac{x}{\log(\frac 1{x})}$ as $x$ goes to $0$. We want to study how much can this decay be slowed while preserving the property highlighted above. Therefore, we now consider for fixed $C>0,D>1$, the following recurrence $$ m_{k+1} = C D^k \frac{m_k}{\log(\frac 1{m_k})} $$

Question : Does there exist an $\epsilon>0$ such that for $0<m_0 < \epsilon$, we have $\lim_{k \to \infty} m_k = 0$?

The trouble here is not only with the non-autonomous nature of the RHS but also the fact that iteration is failing for me. For example, $$ m_{k+1} = CD^k \frac{m_{k}}{\log(\frac 1{m_k})} = CD^k \frac{CD^{k-1} \frac{m_{k-1}}{\log(\frac 1{m_{k-1}})}}{\log\left(\frac{\log(\frac 1{m_{k-1}})}{CD^{k-1}m_{k-1}}\right)} $$

and instantly, one sees logarithms being composed and this leads to analysis which has been long-winded and unfruitful.

I would like to understand what kind of technique might go behind the analysis of such a simple-looking problem once iteration doesn't work. Note that techniques such as fixed point theorems are ruled out because of the non-autonomous nature of the RHS. I'll drop some Python code for those wanting to simulate this iteration.

from math import *
def iterrec(c,d,n,mzero) :
    mk = mzero
    for k in range(n) :
        mk = c(dk)  (mk/log(1/mk))
        print(mk)
iterrec(2,1.01,100,0.05)

In the above call, $C=2,D= 1.01, m_0 = 0.05$ and it shows the values $m_0$ to $m_{100}$. This seems to converge to $0$. Change $C$ to $5$ and it doesn't work : quickly enough, it becomes undefined when some $m_k>1$. However, with $C=5,D=1.01,m_0 = 0.005$ it works again. This suggests that convergence does occur, but I'd like a rigorous analytical proof.

Answer 1

No, it doesn't exist. Let $A=\log C$, $B=\log D$ and $a_k=-\log m_k$, then we get $$a_{k+1}=a_k+\log a_k-A-kB.$$ Suppose that $A+kB\geqslant 0$ when $k\geqslant K$ (as we may since $B>0$). Then we have $a_k\leqslant b_k$ for $k\geqslant K$, where $b_K=a_K$ and $b_{k+1}=b_k+\log b_k$. Now if $b_K>1$ then $b_k\asymp k\log k$ as $k\to\infty$. Thus $\log a_k-A-kB\to-\infty$, so that eventually $a_{k+1}<a_k$ (and even $a_k<0$, i.e. $m_k>1$).