<s>[INST] Check the convergence of $a_{n+1}=\sqrt{a_n+\frac{4}{a_n}}$ where $a_1=4$ Check the convergence of $a_{n+1}=\sqrt{a_n+\frac{4}{a_n}}$ where $a_1=4$. If it converges, find its limit.
I tried to prove that the sequence is monotonically decreasing and bounded by 2, but I failed. Even though, I know that if there exists a limit, than it should satisfy the equation: $a^3-a^2=4$, and its solution is $a=2$.
 [/INST] Let $x$ be a positive real. Then by arithmetic-geometric means inequality we have
$$
\frac{x+4/x}{2} \ge \sqrt{x\cdot \frac{4}{x}}=2,
$$
with equality iff $x=4/x$, i.e. $x=2$. 
Then, we miss only to check that the function is monotonically decreasing starting from $a_1>2$. Take $x>2$, then
$$
x^2>x+\frac{4}{x} \text{ if and only if }x^3-x^2-4>0.
$$
Its first derivative is $3x^2-2x=3x(x-2/3)>0$, hence it is increasing for $x>2$ Therefore $a_{n+1}>a_n$ for all $n$.
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