VMO

Firstly, by induction, one can prove that $$a_n>1,\forall n>1$$ because the function $u(x)=2^{x+1}-x-2$ is increasing on $(1,+\infty )$. It implies that $$a_{n+1}=3-\frac{a_n+2}{2^{a_n}}<3,\forall n\ge 1.$$ Next, we will prove that $(a_n)$ is an increasing sequence. Considering the function $f(x)=3-\frac{2+x}{2^x}$ where $1<x<3$. We have $$f^{\prime }(x)=\frac{\ln 4+x\ln 2-1}{2^x}>0,$$ so $f(x)$ is increasing on $(1,3)$. Moreover $a_2=\frac{3}{2}>a_1$ so $(a_n)$ is an increasing sequence. Note that $(a_n)$ is bounded above by $3$ so $(a_n)$ has a finite limit. Let $L\in (1,3)$ be that limit. Letting $n$ tends to infinity, we have $L=3-\frac{L+2}{2^L}$. We will prove the equation $$x=3-\frac{x+2}{2^x}(1)$$ has a unique solution on $(1,3)$. Indeed, consider the function $g(x)=3-\frac{x+2}{2^x}-x$ where $x\in (1,3)$, we have $$g^{\prime }(x)=\frac{\ln 4+x\ln 2-1}{2^x}-1=\frac{\ln 4+x\ln 2-1-2^x}{2^x},$$ Setting $h(x)=\ln 4+x\ln 2-1-2^x$, for $x\in (1,3)$ then $$h^{\prime }(x)=\ln 2(1-2^x)<0,\forall x\in (1,3),$$ so $h(x)$ is a decreasing function on $(1,3)$. Thus, $h(x)<h(1)=\ln 8-3<0$, or $\ln 4+x\ln 2-1-2^x<0$ for all $x\in (1,3)$. Thus, the function $g(x)$ is decreasing on $(1,3)$ and the equation $g(x)=0$ has no more than one solution. Furthermore, $g(2)=0$ so $x=2$ is the unique solution of (1) which means $L=2$ is the limit of the given sequence. $\square$