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Represent $\sqrt{2}$ in the form $$\sqrt{2}=1+\frac{8}{A_1+\displaystyle\frac{8}{A_2+\displaystyle\frac{8}{A_3+\ddots}}},$$ where $A_n$ is a positive integer and $A_n \geq 8$ for all $n$. So we have $A_1=19$, $A_2=25$, $A_3=15$ and so on. Is the sequence $\left(A_n\right)_{n=1}^\infty$ bounded? How to proof? Thank you.

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Why?????????????????????????????? – Will Jagy Jan 15 '14 at 20:49
I would think not. – Balarka Sen Jan 15 '14 at 20:58
Here are some more terms: – Matthew Conroy Jan 15 '14 at 21:22
Help me please. – kong Jan 18 '14 at 14:32

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