The convergence of $\sqrt {2+\sqrt {2+\sqrt {2+\ldots}}}$ [duplicate]

Solution 1:

Suppose $x\lt 2$

Consider $y=\sqrt {2+x}$ so that $y^2=2+x\lt4$ and $y$ being positive we have $y\lt 2$

That ought to enable you to prove a bound.

Solution 2:

Agreeing with what the others have said, I would like to add that there is in fact a simple explicit formula for the terms of this sequence:

$$ \sqrt{2 + \sqrt{2 + \cdots + \sqrt{2 + \sqrt{\vphantom{\large A}2\,}\,}\,}\,}\ =\ 2\cos\left(\vphantom{\Large A}\pi \over 2^{n + 1}\right) $$

where the square root sign appears $n$-times. In particular, the sequence clearly converges to $2$.

Proof:

For $n=1$, the claim is true, since $\cos(\pi/4)=\sqrt{2}/2$. By the half-angle formula $$2\cos(x/2)=\sqrt{2+2\cos(x)}$$ Therefore $$\sqrt{2+\sqrt{2+\cdots+\sqrt{2+\sqrt{2}}}}=\sqrt{2+2\cos\left(\frac{\pi}{2^n}\right)}=2\cos\left(\frac{\pi}{2^{n+1}}\right)$$ where in the left square root expressions there are $n$ square roots and in the first equality we have used the induction hypothesis that the claim holds for $n-1$.

(From my answer to this question.)

Solution 3:

Let $l=\sqrt {2+\sqrt {2+\sqrt {2+\ldots}}}$, then $l=\sqrt {2+l}$.

From this we have $l^2-l-2=0$

Can you solve it from here? It has to be bounded if there's a finite solution in the limit as $n\to\infty$ and the sequence is monotone.

Solution 4:

Let $a_n = \underbrace{\sqrt{2 + \sqrt{2 + \ldots + \sqrt{2}}}}_{n\;\text{terms}}$ and $\epsilon_n = 2-a_n$. It is clear $$a_1 = \sqrt{2} \quad\implies\quad \epsilon_1 = 2-\sqrt{2} \in(0,1).$$ Notice $$\epsilon_{n+1} = 2 - \sqrt{2+a_n} = 2 - \sqrt{4-\epsilon_n} = \frac{\epsilon_n}{2 + \sqrt{4-\epsilon_n}} \quad\implies\quad \epsilon_{n+1} \in \left(0,\frac{\epsilon_n}{2}\right) $$ We find $\displaystyle\;|\epsilon_n| < \frac{\epsilon_1}{2^{n-1}} < \frac{1}{2^{n-1}} \to 0\;$ as $n \to \infty$. As a result, $\;a_n \to 2\;$ as $\;n \to \infty$.

Solution 5:

As written it's not clear this is a sequence at all but I'm assuming your sequence is

$\sqrt{2}$, $\sqrt{2 + \sqrt{2}}$, $\sqrt{2 + \sqrt{2+ \sqrt{2}}}$, $\sqrt{2 + \sqrt{2+ \sqrt{2+\sqrt{2}}}}$, $\sqrt {2+\sqrt {2+\sqrt {2+ \sqrt{2 +\ldots}}}}$.

Then it should be obvious that each entry in the sequence is positive and the sequence increases each time.

Given this the final entry in the infinite sequence would be $x = \sqrt{2 + x}$ which we can easily solve.

$$\begin{align} x &= \sqrt{2+x} \\ x^2 &= 2 + x\\ x^2 - x -2 &= 0\\ x &= \dfrac{1 \pm \sqrt{1+8}}{2} \\ x &= \dfrac{4}{2} = 2 \end{align}$$

The sequence has a lower bound of $\sqrt{2}$ and an upper bound of $2$ increasing monotonically.