Find limit of recursive sequence $a_{n+1} =\sqrt{2+ \sqrt{a_n} }.$

Question

Let $a_n$ be the sequence in $\mathbb{R}$ given by:

$$ a_1 = \sqrt{2}, \; \; a_{n+1} =\sqrt{2+ \sqrt{a_n} }. $$

Show that this sequence converges and find the limit.

I have already proven that this sequence is increasing and bounded by $2$, so the limit exists. Now, when I'm trying to find the limit, say $L$, I get that $L$ satisfies the equation: $$L^2 -2 - \sqrt{L}=0, $$ but I don't actually know how to solve this equation, so I would like to know if there is an easier way to find the limit, and in case I must solve the last equation, how can I solve it?

Thanks in advance!

Answer 1

Let $M = \sqrt{L}$ then $$M^4-M-2=0,$$ with $M=-1$ being the obvious solution. You need to divide the LHS by $M+1$ and you get a cubic which is solvable exactly with Cardano's Formula.

If you are lazy, just plug into Wolfram Alpha to get $$ M = \frac13 \left(1 - 2 \left(\frac{2}{47 + 3 \sqrt{249}}\right)^{1/3} + \left(\frac{47 + 3 \sqrt{249}}{2}\right)^{1/3}\right) $$

Answer 2

You are on the right track. The square root in your equation for $L$ is awkward. So let's get rid of it by writing $L=x^2$; we can recover $L$ from $x$ at the end by squaring. The equation to solve becomes$$f(x):=x^4-x-2=0.$$The positive solution can be written down, after much labour, as a complicated expression involving nested radicals. However, if you want to know what it is numerically, the easy way is to use Newton–Raphson iteration. This method has the lovely property of quadratic convergence, and only a few iterations are needed to reach the limit of precision of a hand calculator. The basic formula is $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}\quad(n=0,1,...),$$with a given starting value of $x_0$. In the present case, $f'(x)=4x^3-1$, and the formula simplifies to$$x_{n+1}=\frac{3x_n^4+2}{4x_n^3-1}.$$Let's start with $x_0=\frac43$ as a reasonably simple first estimate of the root (which is actually an underestimate by about $1.5$%). The formula gives $x_1=310/229$, which is out by less than $4$ parts in $10,000$. The error in $x_2$ is below $4$ parts in $10,000,000$, and the error in $x_3$ is smaller than an ordinary calculator will show.

Answer 3

Look for a fixed point such that $x = a_{n+1} = a_n$:

\begin{align*} x &= \sqrt{2 + \sqrt{x}} \\ x^4 - 4x^2 - x + 4 &= 0 \\ (x-1)(x^3 + x^2 - 3x - 4) &= 0 \\ \end{align*}

The $x=1$ root is out of range of the sequence, so the only remaining fixed point is the root to $x^3 + x^2 - 3x - 4 = 0$ which is approximately $x \approx 1.8311772$.