# Evaluating an infinite square root

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.

• @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). Jul 26, 2014 at 0:16
• @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. Jul 26, 2014 at 0:28
• 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", .... Jul 26, 2014 at 0:55

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 $285$.