Simplifying $\sqrt{5+2\sqrt{5+2\sqrt{5+2\sqrt {5 +\cdots}}}}$ How to simplify the expression:
$$\sqrt{5+2\sqrt{5+2\sqrt{5+2\sqrt{\cdots}}}}.$$
If I could at least know what kind of reference there is that would explain these type of expressions that would be very helpful.
Thank you.
 A: Put
$$x_0:=0,\quad x_1:=\sqrt{5},\quad x_2:=\sqrt{5+2\sqrt{5}},\quad x_3:=\sqrt{5+2\sqrt{5+2\sqrt{5}}}\ ,$$
and so on, which amounts to
$$x_0:=0,\qquad x_{n+1}:=\sqrt{5+2x_n}\quad(n\geq0)\ .$$
Then
$$x_{n+1}-x_n=\sqrt{5+2x_n}-\sqrt{5+2x_{n-1}}={2(x_n-x_{n-1}) \over \sqrt{5+2x_n}+\sqrt{5+2x_{n-1}}}\ .$$
As $x_1-x_0>0$ this shows that the sequence $(x_n)_{n\geq0}$ is momotonically increasing. 
Furthermore $0\leq x_0<4$, and for any $n\geq0$ the statement $0\leq x_n< 4$  implies $$0\leq x_{n+1}<\sqrt{5+2\cdot 4}<4\ .$$ This shows that our sequence is as well bounded, so it has a limit $\xi\in[0,4]\ $. This limit satisfies the equation $x=\sqrt{5+2x}$ and  is therefore given by $\xi=1+\sqrt{6}\doteq 3.45$.
A: We should really make the problem precise, and prove convergence. But this is the GRE, we manipulate. Let $x$ be the number. Then $x^2-5=2x$. Our number is the positive root of the quadratic.
A: Let $x = 2\sqrt{5+2\sqrt{5+2\sqrt{5+2\sqrt{...}}}}$.  Then (if this converges) $x = 2\sqrt{5+x}$.  Solving, $x = 2(1+\sqrt6)$, so the answer to your original question is $1+\sqrt{6}$
A: Let $x=\sqrt{5+2\sqrt{5+2\sqrt{5+2\sqrt{\dots}}}}$. 
Then $x^2=5+2\sqrt{5+2\sqrt{5+\sqrt{5+2\sqrt{\dots}}}}$.
So, $x^2-5=2\sqrt{5+2\sqrt{5+\sqrt{5+2\sqrt{\dots}}}}$
Remember that $x=\sqrt{5+2\sqrt{5+2\sqrt{5+2\sqrt{\dots}}}}$
So, $2\sqrt{5+2\sqrt{5+\sqrt{5+2\sqrt{\dots}}}}$. So, $2\sqrt{5+2\sqrt{5+\sqrt{5+2\sqrt{\dots}}}}=2x$
This automatically makes our equation to be $x^2-5=2x$. Just solve the quadratic equation, and take the positive root. Why only positive? Think about it: how can $\sqrt{5+2\sqrt{5+2\sqrt{5+2\sqrt{\dots}}}}$ equal a negative number? It must be positive! So that's why we only take the positive root.
$$x^2-2x-5=0$$
$$x=\dfrac{2\pm \sqrt{(-2)^2-4(1)(-5)}}{2(1)}$$
$$x=\dfrac{2\pm \sqrt{4+20}}{2}$$
$$x=\dfrac{2\pm \sqrt{24}}{2}$$
$$x=\dfrac{2\pm 2\sqrt{6}}{2}$$
$$x=1\pm \sqrt{6}$$
Since $1-\sqrt{6}$ equals a negative number, we reject that root. So, $x=1+\sqrt{6}$, which is also the answer to the original problem.
Answer:
$$\displaystyle \boxed{\sqrt{5+2\sqrt{5+2\sqrt{5+2\sqrt{\dots}}}}=1+\sqrt{6}}$$
