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In my answer to the recent question Nested Square Roots, @GEdgar correctly raised the issue that the proof is incomplete unless I show that the intermediate expressions do converge to a (finite) limit. One such quantity was the nested radical $ \sqrt{1 + \sqrt{1+\sqrt{1 + \sqrt{1 + \cdots}}}} \tag{1} $

To assign a value $Y$ to such an expression, I proposed the following definition. Define the sequence $\{ y_n \}$ by: $ y_1 = \sqrt{1}, y_{n+1} = \sqrt{1+y_n}. $ Then we say that this expression evaluates to $Y$ if the sequence $y_n$ converges to $Y$.

For the expression (1), I could show that the $y_n$ converges to $\phi = (\sqrt{5}+1)/2$. (To give more details, I showed, by induction, that $y_n$ increases monotonically and is bounded by $\phi$, so that it has a limit $Y < \infty$. Furthermore, this limit must satisfy $Y = \sqrt{1+Y}$.) Hence we could safely say (1) evaluates to $\phi$, and all seems to be good.

My trouble. Let us now test my proposed idea with a more general expression of the form $\sqrt{a_1 + \sqrt{a_2 + \sqrt{a_3 + \sqrt{a_4+\cdots}}}} \tag{2}$ (Note that the linked question involves one such expression, with $a_n = 5^{2^n}$.) How do we decide if this expression converges? Mimicking the above definition, we can write: $ y_1 = \sqrt{a_1}, y_{n+1} = \sqrt{a_{n+1}+y_n}. $ However, unrolling this definition, one get the sequence $ \sqrt{a_1}, \sqrt{a_{2}+ \sqrt{a_1}}, \sqrt{a_3 + \sqrt{a_2 + \sqrt{a_1}}}, \sqrt{a_4+\sqrt{a_3 + \sqrt{a_2 + \sqrt{a_1}}}}, \ldots $ but this seems little to do with the expression (2) that we started with.

I could not come up with any satisfactory ways to resolve the issue. So, my question is:

How do I rigorously define when an expression of the form (2) converges, and also assign a value to it when it does converge?

Thanks.

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    Isn't this equivalent to the inverse question, whether the sequence $\small y^2,(y^2-a_1)^2,((y^2-a_1)^2-a_2)^2,...$ increases unboundedly for some initial value *y* and the given set of $\small a_k$? Perhaps this is easier to prove.2011-09-01

2 Answers 2

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I would understand it by analogy with continued fractions and look for a limit of $\sqrt{a_1}$, $\sqrt{a_1+\sqrt{a_2}}$, $\sqrt{a_1+\sqrt{a_2+\sqrt{a_3}}}$, ..., $\sqrt{a_1+\sqrt{a_2 \cdots + \sqrt{a_n}}}$, ...

Each of these is not simply derivable from the previous one, but neither are continued fraction approximants.

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    Even though @Bill's answer has been incredibly useful (in fact, it immediately shows that all the expressions involved in the linked question do converge), I think Henning's answer explicitly addresses my definition-trouble. So I will accept this one.2011-09-01
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Vijayaraghavan proved that a sufficient criterion for the convergence of the following sequence $\ \sqrt{a_1 + \sqrt{a_2 +\:\cdots\: +\sqrt{a_n}}}\ \ $ is that $\displaystyle\ \ {\overline {\lim_{n\to\infty}}}\ \frac{\log\:{a_n}}{2^n}\: < \:\infty\:.\: $

For references see see this 1935 Monthly article, Herschfeld: On infinite radicals, and Raoa and Berghe: On Ramanujan's nested roots expansion 1, 2005 and see this prior answer.

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    The manipulations in the previous question even suggest a proof why this criterion works. If we let $K$ stand for the limsup in the criterion, then dividing each approximant by $e^K$ (and distributing this division down through the radical chain) gives the approximants for a different nested radical in which the sequence of $a_i$'s is _bounded_, say by $M$. Then we can bound the approximants from above by the approximants to $\sqrt{M+\sqrt{M+\sqrt{M+\cdots}}}$, which converge by iterating $x\mapsto\sqrt{M+x}$. The approximant sequence is clearly non-decreasing, so this establishes convergence.2011-09-02