# 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 $a=2$.

• So you have already proved that the sequence is bounded? – 3x89g2 May 2 '15 at 8:22

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 $n$.
Think of this in terms of $f(x)=\sqrt{x+\frac{4}{x}}$ http://www.wolframalpha.com/input/?i=sqrt%28x+%2B+4%2Fx%29. See the "Derivative" section, which shows $f(x)$ being ascending for $x>2$.
Then construct its Lamere Ladder. Specifically look at the intersection (one in this case) of $f(x)$ with $g(x)=x$. That's the answer (it depends on the starting point of course, don't take it as a rule ;)).