A bound for the products of a bounded sequence $\{x_{k}\}$ with $\lim_{t\to \infty}\prod_{k=0}^{t}(1-x_{k})=0$

Question

Let $\{x_{k}\}$ be a bounded sequence with $x_{k}\neq 1$, and if \begin{align*} \lim_{t\to \infty}\prod_{k=0}^{t}(1-x_{k})=0. \end{align*} The question is can we infer \begin{align*} \prod_{k=0}^{t}x_{k}< \alpha |x_{0}|2^{t} \end{align*} for some $\alpha>0$？

Here is my initial thought: In the special case where $x_{k}$ is a constant, then $x_{k}\in (0, 2)$, and the inequality holds.

Answer 1

Even excluding the case where any $x$ is $1$ (and the case $x_0=0$ which I assume you also intended to exclude), still the answer is negative. Just have $x_k$ alternate between 2.9 and 1.5, for example. The product of $x$ grows exponentially with base $\sqrt{2.9\cdot1.5}=\sqrt{4.35}>2$ and the product of any two consecutive $1-x$ is $0.95<1$ so that product is $0$.

I should also note that the $|x_0|$ in your proposed conclusion is a pleonasm since $\alpha >0$ is arbitrary and since you need to exclude $x_0=0$ anyway.

Answer 2

You cannot infer that. For example, let $x_0=1$ and $x_j=3$ for all other $j$.