7
$\begingroup$

I would like to know if this sequence converges $\sqrt {2+\sqrt {2+\sqrt {2+\ldots}}}$.

I know this sequence is increasing monotone, but I couldn't prove it's bounded.

Thanks

$\endgroup$

marked as duplicate by Aryabhata, Rebecca J. Stones, Namaste, Joe Johnson 126, M Turgeon Jun 27 '14 at 12:48

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Is this a sequence ? I don't think so ! May be you want to calculate the value of this number. Am I right ? $\endgroup$ – Debashish Jun 27 '14 at 9:24
  • $\begingroup$ Or do you mean the following sequence : $\sqrt{2},\sqrt{2+\sqrt{2}},\sqrt{2+\sqrt{2+\sqrt{2}}}.\ldots$ $\endgroup$ – Debashish Jun 27 '14 at 9:27
  • $\begingroup$ @Debashish the sequence is $a_1=\sqrt 2, a_2=\sqrt{2+\sqrt 2}$,etc. $\endgroup$ – user42912 Jun 27 '14 at 9:28
  • $\begingroup$ @Debashish exactly $\endgroup$ – user42912 Jun 27 '14 at 9:28
  • $\begingroup$ @Debashish It's not uncommon for analysts not to be clear in what they write. The sequence at hand is the one defined by $x_1=\sqrt 2$ and $x_{n+1}=\sqrt{2+x_n}$. $\endgroup$ – Git Gud Jun 27 '14 at 9:28
9
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ What's your $x$? if your $x$ is equal to $\sqrt {2+\sqrt {2+\sqrt {2+\ldots}}}$, then why you're assuming $x\lt 2$? $\endgroup$ – user42912 Jun 27 '14 at 9:35
  • 2
    $\begingroup$ @user42912 If $x_1=\sqrt 2$ and $x_n=\sqrt {2+x_{n-1}}$ for $n\gt 1$ it is a simple induction to show that $x_n\lt 2$ using this method. You just need to start with $\sqrt 2 \lt 2$. $\endgroup$ – Mark Bennet Jun 27 '14 at 9:37
  • 2
    $\begingroup$ Thank you, maybe it should be interesting to add this commentary to your answer just for clarify the future users. $\endgroup$ – user42912 Jun 27 '14 at 9:41
9
$\begingroup$

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.)

$\endgroup$
  • $\begingroup$ Nice one.${{}}$ $\endgroup$ – Git Gud Jun 27 '14 at 10:09
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ you said "It has to be bounded if there's a finite solution and is monotone." why? $\endgroup$ – user42912 Jun 27 '14 at 9:26
  • $\begingroup$ Sorry, edited it for clarity, I shouldn't be doing math at 2:30am. $\endgroup$ – Silynn Jun 27 '14 at 9:27
  • 1
    $\begingroup$ I don't get hows this answers the question (even partially). The OP wants to prove the sequence is bounded. How does your answer help? $\endgroup$ – Git Gud Jun 27 '14 at 9:30
2
$\begingroup$

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 $\;n \to \infty$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.