Convergence of slowly decreasing sequences

Question

Let $A$ be a large number, say $A=100$, and $a$ be a small number, say $a=0.01$.

Let $x_1=1$ and define $x_{n+1}-x_n=-a \exp(-A/x_n)$. Is it true that $x_n\to 0$?

We know that $x_n$ is strictly decreasing. But I do not know how to show that $x_n>0$ for all $n$, since mathematical induction fails here.

Note that $x=0$ is a fixed point for the iteration.

Answer 1

Hint:   $\,f(x) = x - a e^{-A/x} \gt x - e^{-1/x} \gt 0\,$ on $\,\Bbb R^+\,$ for $\,0 \lt a \lt 1 \lt A\,$. Therefore $\,x_n \gt 0\,$ $\,\implies x_{n+1}=f(x_n) \gt 0\,$.

Answer 2

The dynamical system $(\mathbb{R}^+,h)$, where $h(x) = x - a e^{-\frac{A}{x}}$ is well-defined if and only if $ h: \mathbb{R}^+ \rightarrow \mathbb{R}^+$, i.e. if for $x>0$, $h(x)>0$.

You can show that this inequality is equivalent to $\frac{e^A}{a} x e^{\frac{1}{x}} > 1$. The LHS function takes on a minimum at $x = 1$, where the value is $\frac{e^{A+1}}{a}$. Hence you need a and A to satisfy the condition $\frac{e^{A+1}}{a} > 1$.

Further, if A is positive, you have a monotone decreasing sequence $x_n = h^n(1)$ that is bounded below. In that case $x_n \rightarrow 0$.

The values you specified satisfy these conditions.