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Alright, so I have a question on a little open-book challenge-test thingy that deals with repeating square roots, in a form as follows...


Repeated 2012 times (2012 total square roots)

Over and over and over again. I am newish to TeX, so I am not exactly sure how to model it the way it shows up on paper. It looks sorta like:

$$s_n = \sqrt{n+\sqrt{n+\sqrt{n+\cdots}}}$$

How is something like this simplified?

Working it out logically (I am a highschool freshman, mind you), I get something like this for my example: $3-\frac{1}{6^{2011}}$

Is this correct? It seems like I could use some sort of limit to prove this, but I have not officially gone through anything beyond Geometry. Now, I do own bits and pieces of knowledge when it comes to calculus and such, but not enough to count on with this sorta thing ;)

EDIT|IMPORTANT: This is what I need to prove: $3 > \sqrt{6+\sqrt{6+\sqrt{6+\cdots}}} > 3-\frac{1}{5^{2011}}$

Where the \cdots means however many more square roots are needed to make a total of 2012

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Also, I have no clue what to tag this with... Any ideas? – fr00ty_l00ps Nov 19 '12 at 15:59
Your tag is just fine. If you put +\cdots after your last $6$ inside the braces, you'll get the same kind of format as your book has. – Cameron Buie Nov 19 '12 at 16:06
Thank you, I appreciate that :) – fr00ty_l00ps Nov 19 '12 at 16:10
Related: (See the latter half of the answer, where similar inequalities are proven for the general recurrence). – Aryabhata Apr 24 '13 at 8:38
up vote 2 down vote accepted

If you start with $$x = \sqrt{6+\sqrt{6+\sqrt{6+\cdots}}}$$ with an infinite number of terms then $$x=\sqrt{6+x}$$ which you can solve. Depending on how you do it, perhaps by squaring both sides, this could give two potential solutions and you need to satisfy youself about which, if any, is correct.

With $2012$ sixes I would have thought you could only get an empirical result, perhaps something close to $$3-\frac{3.36566}{6^{2012}}$$ as suggested with seven to thirteen sixes.

With $f(n) = \sqrt{6 +f(n-1)}$ starting with $f(0)=0$, you might try to prove something like $$3-6^{n-1} \lt f(n) \lt 3$$ for $n \ge 1$ by induction: it may not be easy, as I think $\sqrt{6+3-6^{-(n-1)}} \lt 3 - 6^{-n}$.

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And then anything to a negative power is something like $\frac{1}{n^{x+1}}$, where $x$ is negative, right? – fr00ty_l00ps Nov 19 '12 at 16:52
@CodeAdmiral: Yes: $x^{-n} = \dfrac{1}{x^n}$, either by definition, or since multiplying each by $x^n$ gives $x^0=1$, at least for positive $x$. – Henry Nov 19 '12 at 17:00

Your edit suggests using induction. Define $f(0)=\sqrt 6, f(n+1)=\sqrt{6+f(n)}$ We would like to prove $f(n)\gt 3-\frac 1{5^n}$. Certainly this is true for $n=0$, as it asserts $\sqrt 6 \gt 2$. Now assume it is true for $k$, then $f(k+1)=\sqrt {6+f(k)}\gt \sqrt{9-\frac 1{5^k}}=3\sqrt{1-\frac1{9\cdot 5^k}}\gt 3(1-\frac1{2\cdot 9\cdot 5^k})\gt3-\frac1{ 5^{k+1}}$

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