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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,

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marked as duplicate by Claude Leibovici calculus Jun 19 '15 at 8:31

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.

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    $\begingroup$ 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. $\endgroup$ – JimmyK4542 Jun 19 '15 at 3:48
  • $\begingroup$ 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? $\endgroup$ – Chappers Jun 19 '15 at 3:49
  • $\begingroup$ 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: $\sqrt{c + \sqrt{c+\sqrt{c + \ldots}}}$ $\endgroup$ – user230734 Jun 19 '15 at 3:49
  • $\begingroup$ Yes, @chappers - that's what I mean :-) $\endgroup$ – user249229 Jun 19 '15 at 3:53
  • $\begingroup$ Ok, got it - thanks @BolzWeir $\endgroup$ – user249229 Jun 19 '15 at 3:54
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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 $A\big((n(n+1)\big)=n+1$.

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  • $\begingroup$ Nice! Wish they just reopened it $\endgroup$ – Orest Bucicovschi Jun 19 '15 at 6:21
  • $\begingroup$ 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. $\endgroup$ – User001 Jun 19 '15 at 6:30
  • $\begingroup$ ...because then I feel that I am incorrectly assuming that the limit exists and is finite. $\endgroup$ – User001 Jun 19 '15 at 6:33
  • $\begingroup$ 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. $\endgroup$ – User001 Jun 19 '15 at 6:35
  • $\begingroup$ @LebronJames. I cannot say that your approach is wrong (I am not a mùathematician) but it seems to be taking a long way. $\endgroup$ – Claude Leibovici Jun 19 '15 at 6:36

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