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Show that $$\sqrt{2\sqrt{2\sqrt{2\cdots}}}=2$$ $$\sqrt{2}=\mathbf{2}^{1/2}$$ $$\sqrt{2\sqrt{2}}=\mathbf{2}^{1/2+1/2^2}$$ $$\sqrt{2\sqrt{2\sqrt{2}}}=\mathbf{2}^{1/2+1/2^2+1/2^3}$$ Show the limit of $$\mathbf{S}_{n}=\frac{1}{2}+\frac{1}{2^2}+\dotsb+\frac{1}{2^n}=1$$ when $n\to\infty$ $$\textbf{S}_{n}=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^n}$$ $$\Rightarrow \frac{1}{2}\textbf{S}_{n}=\frac{1}{2}(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^n})$$ $$\Rightarrow \frac{1}{2}\textbf{S}_{n}=(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{n+1}})$$ $$\Rightarrow (1)-(2)=\textbf{S}_{n}-\frac{1}{2}\textbf{S}_{n}=\frac{1}{2}-\frac{1}{2^{n+1}}$$ $$\Rightarrow \textbf{S}_{n}(1-\frac{1}{2})=\frac{1}{2}-\frac{1}{2^{n+1}}$$ $$\Rightarrow \frac{1}{2^{n+1}}\rightarrow\textbf{0}\quad\textit{when n}\rightarrow\infty$$ $$\Rightarrow \textbf{S}_{n}\rightarrow\textbf{1}\quad\textit{when n}\rightarrow\infty$$ $$\Rightarrow \lim_{n \to \infty}\textbf{2}^{\textbf{S}_{n}}=2\quad\textit{when n}\rightarrow\infty$$

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    $\begingroup$ See this question: math.stackexchange.com/questions/589288/… $\endgroup$ – 0912 Dec 10 '13 at 7:15
  • $\begingroup$ I started reformatting this but it's too big a job. Please don't embed your text in MathJax formulas. That's not how things are done here. $\endgroup$ – dfeuer Dec 10 '13 at 7:32
  • $\begingroup$ +1 for showing your efforts. The logic in your proof is fine. However, in some steps, the $\implies$ symbol has been misused. (e.g in the line "$\implies (1) = (2) \ldots$" and the line "$\implies \frac{1}{2^{n+1}} \to 0 \text{ when } n \to \infty$" ). $\endgroup$ – achille hui Dec 10 '13 at 7:34
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Let $x_1 = \sqrt{2}$ and define $x_{n+1} = \sqrt{2 x_n}$, then it suffices to show that $\lim_n x_n = 2$. In order to achieve this goal show that the sequence $x_n$ is monotonically increasing and bounded above (I will leave this for you to do).

Then the limit exists so let $x = \lim x_n$. Then, using $x_{n+1} = \sqrt{2 x_n}$ and the continuity of the square root function, we get that $x = \sqrt{2 x} \implies x=2$.

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  • $\begingroup$ @Mustafe Said. $x^2 = \sqrt{2 x} \implies x=2$. Typo, I guess (and hope !). Cheers. $\endgroup$ – Claude Leibovici Dec 10 '13 at 7:44
  • $\begingroup$ I corrected it, thank you Claude. $\endgroup$ – Mustafa Said Dec 10 '13 at 7:46
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Let S be your general term for a large value of n. If your square it you have S^2 = 2 S, then S = 2.

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It's more simple to prove that, let $\sqrt{2\sqrt{2\sqrt{2}...}}=t$. Then, $t^2=2t \rightarrow t=2$.

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  • $\begingroup$ $$\text{You assume}\quad\sqrt{2\sqrt{2\sqrt{2...}}}=\sqrt{2\sqrt{2…}}\quad\text{it is true for final term}$$ $\endgroup$ – bsdshell Dec 10 '13 at 9:09
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$$T=\sqrt{2\sqrt{2\sqrt{2}}}...\\ \frac{T^2}{2}=T\\$$ T is a nonzero real number so:$$ T=2$$

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Your proof is correct!

Here is a different proof.
First note that the sequence, let's call it $(a_n)_{n\in\mathbb N}$, can be defined recursively as $a_1=\sqrt{2}$ and $a_{n+1}=\sqrt{2a_n}$. Then you can show that it is increasing and bounded above by 2, therefore it converges.
Its limit, $\ell$, must satisfy $\ell=\sqrt{2\ell}$, and therefore $\ell=2$.

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