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Possible Duplicate:
Limit of the nested radical $\sqrt{7+\sqrt{7+\sqrt{7+\cdots}}}$

how can one solve for $x$, $x =\sqrt[]{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2\cdots }}}}}}$

we know, if $x=\sqrt[]{2+\sqrt{2}}$, then, $x^2=2+\sqrt{2}$

now, if $x=\sqrt[]{2+\sqrt{2}}$, then, $(x-\sqrt{2})(x+\sqrt{2})=\sqrt{2}$

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    This may be relevant: http://math.stackexchange.com/questions/179981/how-can-one-express-sqrt2-sqrt2-without-using-the-square-root-of-a-squar/180000#1800002012-08-25
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    What can you say about $x^2$?2012-08-25
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    See also: [this thread](http://math.stackexchange.com/q/61048/5363), [this thread](http://math.stackexchange.com/q/132926/5363), [this thread](http://math.stackexchange.com/questions/115501/on-the-sequence-x-n1-sqrtcx-n) and the links therein.2012-08-25
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    If I wanted to be quirky, I'd say that $x$ is already solved for.2012-08-25
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    Is that expression even well-defined? What does an infinite series of square roots actually mean? Is this some kind of limit as $n\to\infty$? It's easy to show that if this limit exists, then it must equal 2 (see the answers below); but a full solution would have to show that there is some kind of convergence going on.2012-08-25
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    I first saw this in a problem solving seminar for math majors-it's really a trick question because all you have to realize is that no matter how complicated the expression is, since x is an undetermined variable, you can simply assign x to the whole mess on the right side of the first square root. Then high school algebra gets you a quadratic equation and the rest is history. : )2012-08-25
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    @David That's a very good point you've made,but if we replace the square roots with rational exponents, the resulting infinite sum of (2^(1/(2^n)) converges to 2 since if we square both sides and manipulate the sum, we obtain 2 + (2^(1/(2^(n-1))). The right hand sum clearly converges to 0 as n goes to infinity. Simple calculus.2012-08-25
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    Sorry, @Mathemagician1234, but I don't see how your argument proves anything. I think the best way to go is to show that the sequence of expressions with a finite number of square roots is monotonically increasing and bounded above. You need to do something like this to show that the limit is defined.2012-08-25
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    @DavidWallace : replace "the limit defined" with "the limit is classically defined". There are plenty of examples of non bounded series that converge but not in the classical sense.2012-08-25
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    @Arjang *There are plenty of examples of non bounded series that converge but not in the classical sense*... No idea what you are talking about here. Care to elaborate?2012-08-25
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    @did : Just want to remind that classical convergence is not the only game in town.2012-08-25
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    @Arjang Then such a, universal, reminder might not be completely relevant as an answer to David Wallace's comment which, as far as I can see, quite properly reminds some very basic aspects of limits, in the most classical sense of the term, seemingly overlooked by some on this page. In the end, "the limit is defined" is perfect and I would not "replace" it by "the limit is classically defined".2012-08-25
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    @did : that is true.2012-08-25
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    I really don't see how this question is a duplicate of the linked one. While similar proofs work, there are solutions which work for $2$ but not for the linked one..... After all $2 \neq 7$.2013-05-31

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