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How do I evaluate the square root: $$\sqrt{2013+276\sqrt{2027+278\sqrt{2041+280\cdots}}}$$ I have tried creating two arithmetic sequences such that $$a_n = 1999+14n$$

$$b_n = 274+2n$$ so the square root simplifies to $$\sqrt{a_1+b_1\sqrt{a_2+b_2\sqrt{a_3+b_3\cdots}}}$$ But I get stuck there. Any help/hints is greatly appreciated.

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  • $\begingroup$ @ozo: And $c_{n+1}=\frac{c_n^2-a_n}{b_n}$, so you can choose any sufficiently large $c_1$ and extend it to a sequence of $c_i$s that satisfies these relations. That's not really helpful for finding the value sought here (which I assume is the limit of the truncated nested radicals). $\endgroup$ Jul 26, 2014 at 0:16
  • $\begingroup$ @HenningMakholm I don't understand what you mean by truncated, but it's an infinite nested radical, where there's an infinite amount of terms. $\endgroup$ Jul 26, 2014 at 0:28
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    $\begingroup$ Check some of the question listed under "Related" and see whether there's anything you can use there. Or search the site for "nested radical", "nested square root", .... $\endgroup$ Jul 26, 2014 at 0:55

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It just a special case of the Ramanujan nested radical: $$a+n+x=\sqrt{ax+(n+a)^2 +x\sqrt{a(x+n)+(n+a)^2+(x+n) \sqrt{\mathrm{\cdots}}}},$$ with the choices $a=7,x=276,n=2$, that ensure convergence to $285$.

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  • $\begingroup$ How do you prove this identity? $\endgroup$ Jul 26, 2014 at 12:25

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