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$\sqrt{2 + \sqrt{2+\sqrt{2+\sqrt2...)}}}$.

Pretty classic question, I think - and the limit is equal to 2.

But how do I prove this rigorously? An epsilon-delta proof wouldn't work, since I wouldn't know the limit is equal to 2 - the question asks, if the limit exists, compute it. This was for an old analysis exam, not a calculus class, so I feel that I can't just set the above = some number L, and then make algebraic manipulations on both sides of the equation, until I get what I want. We can't assume the limit exists, I think.

Thanks,

Martin Sleziak
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user249229
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    Define a sequence by $a_0 = \sqrt{2}$, and $a_{n+1} = \sqrt{2+a_n}$ for $n \ge 0$. Now, show that $\displaystyle\lim_{n \to \infty}a_n = 2$. This has probably been done on this site before. – JimmyK4542 Jun 19 '15 at 03:48
  • I think the first thing you need to decide is what the symbolic expression you have there actually means. @JimmyK4542 has a reasonable suggestion, but is this what you mean? – Chappers Jun 19 '15 at 03:49
  • Pretty duplicate as well. Every week or so, someone asks the same question. I think that there's a community wiki thing which handles a general limit: $\sqrt{c + \sqrt{c+\sqrt{c + \ldots}}}$ –  Jun 19 '15 at 03:49
  • Yes, @chappers - that's what I mean :-) – user249229 Jun 19 '15 at 03:53
  • Ok, got it - thanks @BolzWeir – user249229 Jun 19 '15 at 03:54
  • Note that $a_{n+1}^2 - 4 = a_n -2$ so $| a_{n+1} -2| = \frac{1}{|a_{n+1}+2|} \cdot |a_n-2|$, hence $\log|a_n-2| \simeq - n \log 4$, the convergence is exponential – orangeskid Jun 19 '15 at 06:27
  • See http://math.stackexchange.com/questions/555778/show-that-sqrt2-sqrt2-sqrt2-converges-to-2, http://math.stackexchange.com/questions/849274/the-convergence-of-sqrt-2-sqrt-2-sqrt-2-ldots or more general http://math.stackexchange.com/questions/115501/sqrtc-sqrtc-sqrtc-cdots-or-the-limit-of-the-sequence-x-n1-sq – Martin Sleziak Jun 19 '15 at 07:53
  • I wanted to vote to close as a duplicate, but I see that it has already been closed and reopened. Maybe somebody should explain why this is not a duplicate of some of the linked questions. – Martin Sleziak Jun 19 '15 at 07:57

1 Answers1

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If $$A=\sqrt{2 + \sqrt{2+\sqrt{2+\sqrt2+\cdots}}}$$ just square and get $$A^2=2+\sqrt{2 + \sqrt{2+\sqrt{2+\sqrt2+\cdots}}}=2+A$$ So $A^2-A-2=0$ and then the solution.

More generaly, if $$A(c)=\sqrt{c + \sqrt{c+\sqrt{c+\sqrt c+\cdots}}}$$ $$A^2(c)=c+A(c)$$ $$A(c)=\frac{1}{2} \left(1+\sqrt{4 c+1}\right)$$

You will find whole numbers for $c=2,6,12,20,30,42,56,72,90,110,132,156,182,\cdots$ that is to say for $c=n(n+1)$ which gives $A\big((n(n+1)\big)=n+1$.

Claude Leibovici
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  • Nice! Wish they just reopened it – orangeskid Jun 19 '15 at 06:21
  • Hi @ClaudeLeibovici, I instead wrote $a_n = e^{(log(\sqrt{(2+a_{n-1}}))}$ and taking the limit on both sides, as n goes to infinity, by the continuity of the exponential and log functions, I can swap the limit with both functions. I arrive at (lim $a_n$)^2 = 2 + lim$a_n$. I now move the right-hand terms over to the left and get a polynomial equation in x = (lim$a_n$). Solving for the zeroes gives me x = 2 or x = -1. Since the sequence is clearly positive, I take x to be 2, and so lim $a_n$ = 2. Do you think any part of my work is wrong? I'm not sure if I can label lim$a_n$ = x. – User001 Jun 19 '15 at 06:30
  • ...because then I feel that I am incorrectly assuming that the limit exists and is finite. – User001 Jun 19 '15 at 06:33
  • Hi @JimmyK4542, I used your hint and my work is summarized above. What do you think? I am just shaky about labeling lim $a_n$ = x and solving the (quadratic) polynomial equation, and taking the positive zero, x = 2. – User001 Jun 19 '15 at 06:35
  • @LebronJames. I cannot say that your approach is wrong (I am not a mùathematician) but it seems to be taking a long way. – Claude Leibovici Jun 19 '15 at 06:36
  • Ok, got it. Thanks so much for your help, @ClaudeLeibovici. – User001 Jun 19 '15 at 06:40
  • @LebronJames. You are very welcome ! Remember : being lazy in maths is not a sin ! When the problem looks very complex for an assignment in a limited time, consider that there is probably a simple way to do it. Cheers :-) – Claude Leibovici Jun 19 '15 at 06:41
  • @ClaudeLeibovici I see you have reopened the question. If it is not a duplicate of some of the linked questions, it would be useful to explain in the comments to the question why it is not a duplicate. Otherwise it is very probable that it will be closed again. – Martin Sleziak Jun 19 '15 at 07:59
  • @MartinSleziak. You are totally right and I apologize. I made that by mistake. Being almost totally blind, this happens to me quite (too) often (clicking the wrong button). Sorry again :-( – Claude Leibovici Jun 19 '15 at 08:31
  • No problem. I am glad that the thing is clarified. (I thought that I might have missed some reason why the two questions should not be duplicates, so I asked first.) – Martin Sleziak Jun 19 '15 at 08:32