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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

J.R.
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user42912
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  • Is this a sequence ? I don't think so ! May be you want to calculate the value of this number. Am I right ? – Debashish Jun 27 '14 at 09:24
  • Or do you mean the following sequence : $\sqrt{2},\sqrt{2+\sqrt{2}},\sqrt{2+\sqrt{2+\sqrt{2}}}.\ldots$ – Debashish Jun 27 '14 at 09:27
  • @Debashish the sequence is $a_1=\sqrt 2, a_2=\sqrt{2+\sqrt 2}$,etc. – user42912 Jun 27 '14 at 09:28
  • @Debashish exactly – user42912 Jun 27 '14 at 09:28
  • @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 $x_{n+1}=\sqrt{2+x_n}$. – Git Gud Jun 27 '14 at 09:28
  • I'm assuming the sequence is $a_n=\sqrt {2+a_{n-1}}$ – Mark Bennet Jun 27 '14 at 09:29
  • @GitGud you don't need the internal square root in your definition. – Mark Bennet Jun 27 '14 at 09:30
  • @GitGud .. Ya thats true. – Debashish Jun 27 '14 at 09:30
  • @MarkBennet Thanks for pointing it out. – Git Gud Jun 27 '14 at 09:30
  • @GitGud I made that mistake in my answer before I straightened it out ... – Mark Bennet Jun 27 '14 at 09:31
  • @Aryabhata I disagree with marking these questions as duplicates. Special cases often allow for much neater proofs. – user1729 Jun 27 '14 at 11:53
  • @user1729PhD: I would like to refer you to: http://meta.math.stackexchange.com/questions/1756/coping-with-abstract-duplicate-questions and http://meta.math.stackexchange.com/questions/1868/list-of-generalizations-of-common-questions. If the asker is looking for a different answer than those that appear in the generalized question, they can always say so in the question, and a closure will not happen. – Aryabhata Jun 27 '14 at 13:27
  • @Aryabhata I am aware of those threads, but I do not think that they are applicable here. Jeff Atwood interprets the sentiment as "old-timers are tired of answering what is essentially the same question in millions of tiny different varations", but I do not think that this is necessarilly applicable here. It is a much reduced question, not a little reduced question. These threads are more for "find the $\gcd$ of $36$ and $15$" vs. "find the $\gcd$ of $100$ and $6$", no?. – user1729 Jun 27 '14 at 15:41
  • @user1729PhD: I disagree. Where do you draw the line? For instance, who knows, one might find an extremely interesting graph theory proof of $\text{gcd}(100, 6) = 2$. Why close that question? I am repeating myself, but if someone is looking for an answer not given already, they can specifically mention that, and the question will not be closed. – Aryabhata Jun 30 '14 at 05:03

5 Answers5

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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.)

Community
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J.R.
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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.

Mark Bennet
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  • What's your $x$? if your $x$ is equal to $\sqrt {2+\sqrt {2+\sqrt {2+\ldots}}}$, then why you're assuming $x\lt 2$? – user42912 Jun 27 '14 at 09:35
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    @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 $\sqrt 2 \lt 2$. – Mark Bennet Jun 27 '14 at 09:37
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    Thank you, maybe it should be interesting to add this commentary to your answer just for clarify the future users. – user42912 Jun 27 '14 at 09:41
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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.

Silynn
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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.

Warren Hill
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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 $\;n \to \infty$.

achille hui
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