Prove that $a_0 = \sqrt{2}, a_{n+1} = \sqrt{2 + \sqrt{a_n}}$ converges and find its limit Prove that following sequence converges and find its limit
$$a_0 = \sqrt{2}$$
$$a_{n+1} = \sqrt{2 + \sqrt{a_n}}$$
My work: I know I have to prove that sequence is bounded and monotonous in order to prove that it converges, but I don't know how to prove both of it. I tried to prove that $a_n$ is bounded above by 2, and that is true, I got that using induction, but I don't know how to prove that $a_{n+1} - a_n > 0$ or $\frac{a_{n+1}}{a_n}>1$. Limit itself is not as big problem as to prove convergence, what I did is that, recurent equality becomes following, where L is limit
$$L = \sqrt{2+ \sqrt{L}}$$
and from that we can easy get
$$ L^2 - 2 = \sqrt{L}$$
where $L$ must be greater than $\sqrt{2}$. That is same as this equation
$$(L - 1)(L^3 + L^2 -3L - 4)=0$$
and because $L$ can not be 1 then, using Cardano's formula I have got that $L$ is approximately $1.83$.
 A: *

*You can use numerical methods to find an approximate solution. The point is that the sequence $a_n$  is indeed convergent. Proving that it is bounded and that it has monotonicity properties is not complicated in this case. For example $a_1>a_0$ then,  by induction, if $a_n>a_{n-1}$, then $$a_{n+1}=\sqrt{2+\sqrt{a_n}}>\sqrt{2+\sqrt{a_{n-1}}}=a_n$$
Boundedness can also be proven by induction.


*The limit $L$, as you pointed out satisfies the quartic equation in your posting. Notice that if  $u=\sqrt{L}$, then $u$ satisfies the quartic equation
$$p(u)=u^4−u−2=0$$
There is only one change of signs in consecutive nonzero coefficients of the quartic polynomial. Thus, there exactly one real positive solution  $u^*$ (Decartes criteria) and $L^*=(u^*)^2$ is the limit of $a_n$.


*Applying Descartes' to $p(-u)=u^4+u-2=0$, you get that there is also exactly one positive solution, which happens to be $u_*=1$. That results in the extra positive solution that you get from the quartic equation in $L$ in your posting.


*One last comment: the Descartes' criteria applied to
$$q(L)=L^3+L^2-3L-4=0$$
also implies that there is only one positive solution to the cubic equation, for there is exactly one change of sign in consecutive nonzero coefficients of $q$.
