Limit of a sequence including an exponential function

Question

Let $\{a_n\}_{n\ge1}$ be a sequence with $\frac{1}{n}\le a_n\le1$ for each $n$. We have the following two inequalities: $$ a_n e^{\displaystyle-\frac{2+\frac{1}{n}}{a_n}}+\frac{1}{n}e^{-2}\ge e^{-5} $$ and $$ a_n e^{\displaystyle-\frac{2+\frac{1}{n}}{a_n-\frac{1}{n}}}-\frac{1}{2n}e^{-2}\le e^{-5}. $$ I would like to show that $a:=\lim_{n\to\infty}a_n$ exists and $a$ satisfies that $$ ae^{-2/a}=e^{-5}. $$ I have no idea how to guarantee the existence of $a$. Please give me some hints. Thanks a lot.

Answer 1

The sequence $(a_n)$ is bounded. If $s$ is an accumulation point of $(a_n)$ then there is a subsequence, which converges to $s$. Try to show that $s>0$. From the given inequalties we get $se^{-2/s}=e^{-5}.$

Next define $f(x):=xe^{-2/x}$ for $x>0$ and show that $f$ is strictly increasing. It is also your task to show that $f((0, \infty))=(0, \infty).$

Therefore the equation $f(x)=e^{-5}$ has a unique solution in $ \mathbb R.$

Consequence: $(a_n)$ is bounded and has only one accumulation point. Hence $(a_n)$ is convergent.