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$\sqrt{1+\sqrt{2+\sqrt{3+\cdots+\sqrt{2006}}}}<2$. I struggled on it, but i did't find any pattern to solve it.

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Does anyone know the limit of $\sqrt{1+\sqrt{2+\sqrt{3+\cdots +\sqrt{n}}}}$ in closed form? –  lhf May 28 '13 at 11:49
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@OlayinkaSF, thanks for the link! –  lhf May 29 '13 at 1:18
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2 Answers

up vote 8 down vote accepted

$$\begin{aligned}\sqrt{1+\sqrt{2+\sqrt{3+\cdots \sqrt{n}}}}&<\sqrt{1+\sqrt{2+\sqrt{2^2+\cdots \sqrt{2^{2^{n-1}}}}}}\\&<\sqrt{1+\sqrt{2+\sqrt{2^2+\cdots }}}\\&=\sqrt{1+\sqrt{2}\cdot\sqrt{1+\sqrt{1+\cdots }}}\\&=\sqrt{1+\sqrt{2}\phi}\\&<2\end{aligned}$$


We can get a tighter bound for the limit by breaking the pattern a little further down the line:

$$\begin{aligned}\sqrt{1+\sqrt{2+\sqrt{3+\cdots \sqrt{n}}}}&<\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{2^2+\cdots}}}}\\&=\sqrt{1+\sqrt{2+\sqrt{3+2\sqrt{1+\cdots }}}}\\&=\sqrt{1+\sqrt{2+\sqrt{4+\sqrt{5}}}}\approx 1.7665398\end{aligned}$$

$$\sqrt{1+\sqrt{2+\sqrt{3+\cdots }}}\approx 1.7579327$$

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Very nice. And a pretty good bound too, because the limit seems to be $1.757932756618\ldots$ and $\sqrt{1+\sqrt{2}\phi} \approx 1.813352037325$. Does anyone know this limit in closed form? –  lhf May 28 '13 at 11:23
    
In $2^{2^n}$, is there a power too much? –  NiftyKitty95 May 28 '13 at 11:32
    
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Idea: You can unwrap like this:

$$\sqrt{1+\sqrt{2+\sqrt{3+\cdots+\sqrt{2006}}}}<2$$

if

$$\sqrt{2+\sqrt{3+\cdots+\sqrt{2006}}}<2^2-1$$

if

$$\sqrt{3+\cdots+\sqrt{2006}}<(2^2-1)^2-2$$

and so on, so we want to show

$$2006 < (((2^2-1)^2-2)^2-\cdots)^2-2005$$


might as well prove it by induction for all $n$ rather than just 2006, so we need to show that

$$n+1 < (((2^2-1)^2-2)^2-\cdots)^2-n$$

implies

$$n+2 < ((((2^2-1)^2-2)^2-\cdots)^2-n)^2-(n+1)$$

but thats just

$$2n+3 < ((((2^2-1)^2-2)^2-\cdots)^2-n)^2$$

which holds since

$$2n+3 < (n+1)^2$$

for all $n>1$.

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