<s>[INST] Prove the convergence of the sequence $a_{1} = 4$, $a_{n + 1} = \frac{a_{n}}{2} + \frac{2}{a_{n}}$, $n = 1, 2, \ldots$ Prove the convergence of the sequence $a_{1} = 4$, $a_{n + 1} = \frac{a_{n}}{2} + \frac{2}{a_{n}}$, $n = 1, 2, \ldots$
I'm pretty sure the way to do it is to show $a_{n} > 2$ for $n = 2, \ldots$ and then maybe use the Monotone Convergence Theorem to show it converges to $2$, but I think this also might be wrong. Can someone please help me with this problem? I don't know how to prove a bound for it.
 [/INST] Note, that $\color{blue}{a= 2}$ is a $\color{blue}{\mbox{fixpoint}}$ of the iteration as 
$$2 = \frac{a}{2} + \frac{2}{a} = 1+1$$
AM-GM shows that for all members of the sequence we have 
$$\frac{a_n}{2} + \frac{2}{a_n} \stackrel{AM-GM}{\geq} 2$$
Now, consider
$$0 \leq \color{blue}{a_{n+1}-2} = \frac{a_n}{2} + \frac{2}{a_n} - 2 = \frac{a_n -2}{2} - \frac{a_n - 2}{a_n} = \left( \frac{1}{2} - \frac{1}{a_n} \right)(a_n - 2) \color{blue}{\stackrel{a_n \geq 2}{\leq} \frac{1}{2} (a_n - 2)}$$
It follows
$$0 \leq a_{n+1}-2 \leq \left( \frac{1}{2}\right)^n(a_1 - 2)\stackrel{n \to \infty}{\longrightarrow} 0 \Rightarrow \color{blue}{(a_n) \mbox{ is convergent and } \lim_{n \to \infty}a_n = 2}$$
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