Question: Let $a_n$ be a positive nonzero integer. Are there any known criteria in order for $$\sqrt{a_1+\sqrt{a_2-\sqrt{a_3+\sqrt{a_4-\sqrt{a_5+\ldots}}}}}$$ to converge ?

The motivation here was the initial question: Should we believe $$\sqrt{2+\sqrt{3-\sqrt{5+\sqrt{7-\sqrt{11+\ldots}}}}}$$ converges ? One could equally be curious about $$\sqrt{2+\sqrt{4-\sqrt{6+\sqrt{8-\sqrt{10+\ldots}}}}}$$ and compare that to the nested radical constant. Also observe $$\sqrt{1+\sqrt{2-\sqrt{3+\sqrt{4-\sqrt{5+\ldots}}}}}$$ appears to be a complex number in stark contrast to the nested radical constant. So I guess we should be concerned knowing if $\sqrt{a_1+\sqrt{a_2-\sqrt{a_3+\sqrt{a_4-\sqrt{a_5+\ldots}}}}}$ is a real number.

Note Vijayaraghavan special case of Herschfeld's theorem on nested radicals. But this does not apply to alternating plus and minus. Click here for Herschfeld's paper on nested radicals

  • $\begingroup$ Just to do a comment, not related with your question: maybe you can to state different problems using different arithmetic functions. That I evoke is that instead of your pattern of signs $+,-,+,\ldots$ inside the nested radicals maybe it is interesting do experiments with a computer with functions as the Möbius function or Liouville function (for examples of $ \left\{ a_n\right\}_{n=1}^\infty $). Isn't required a response of this comment, good day. $\endgroup$ – user243301 Sep 29 '17 at 0:40

A search for "nested radical" brings up this article in mathworld:


It only generally talks about radicals where all the signs are positive, unlike your case.

It mentions a theorem that mignt be relevant:

Herschfeld's convergence theorem:

If $0 < p < 1$ and all $x_i \ge 0$ then $\lim_{n \to \infty} x_0 +(x_1+(x_2+(...x_n^p)^p)^p)^p $ exists if and only if $ (x_n)^{p^n}$ is bounded.

The references are

Herschfeld, A. "On Infinite Radicals." Amer. Math. Monthly 42, 419-429, 1935.

Jones, D. J. "Continued Powers and a Sufficient Condition for Their Convergence." Math. Mag. 68, 387-392, 1995.

Your question involves $p = \frac12$.

  • $\begingroup$ You are right. I misquoted the theorem. Thanks. Upvoted. $\endgroup$ – marty cohen Oct 22 '18 at 19:28

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