VMO

a) For $a=0$, we have $(u_n)$ defined by $$u_1=6,u_{n+1}=2+\sqrt{u_n+4},\forall n\in {\mathbb{N}}^{\ast }.$$ It is clear that $u_n\ge 2$ for all positive integers $n$. On the other hand, we get $u_2<u_1$. By induction, we can point out that $(u_n)$ is decreasing. Hence, $(u_n)$ has a finite limit $l$. By letting $n$ tend to infinity, we get $l=2+\sqrt{l+4}$, thus $l=5$.

b) Firstly, we prove that $(u_n)$ is bounded. Let $n_0$ be a positive integer such that $n_0>a$. Now we choose a positive number $M>10$ such that $M>\max (u_1,u_2,\dots ,u_{n_0})$ which implies $$u_{n_0+1}\le 3+\sqrt{2u_{n_0}+4}\le 3+\sqrt{2M+4}\le M.$$ By induction, we can point out that $u_n\le M$ for all positive integers $n$. Note that $u_n>0$ then $(u_n)$ is bounded. Next, we will prove this statement: $(u_n)$ is a monotone sequence (not necessary from the first term). Clearly, if $(u_n)$ is non-decreasing then the statement is proved. On the other hand, there exists a positive integer $m$ that $u_m>u_{m+1}$. Hence, $$\frac{2m+a}{m}+\sqrt{\frac{m+a}{m}{u}_m+4}>\frac{2m+2+a}{m+1}+\sqrt{\frac{m+1+a}{m+1}{u}_{m+1}+4}.$$ Therefore, $u_{m+1}>u_{m+2}$. By induction, we claim that $(u_n)$ is decreasing from $u_m$ and the statement is proved. Note that we also have pointed out that $(u_n)$ is bounded, it implies that $(u_n)$ has a finite limit.