# $\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}\approx 1.5176001678…$

While on chat, an interesting limit popped out: $$\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}\approx 1.5176001678777188...\lt\phi$$ robjohn determined its value for fifty places, and Inverse Symbolic Calculator yields nothing.

Is there a closed form for this nice limit?

• Great question! (+1) – user 1357113 Jul 30 '13 at 22:09
• What closed form do you expect if the Inverter fails? – Hagen von Eitzen Jul 30 '13 at 22:18
• @Chris'ssis Why? – Did Jul 30 '13 at 22:35
• @Did If I answered the question I'd describe myself and I don't wanna do that, not here, not now. :-) – user 1357113 Jul 30 '13 at 22:51
• @Chris'ssis ?? Whatever. – Did Jul 30 '13 at 22:53

Let $$r_n := 1+\sqrt{1+\sqrt{1+\sqrt{\ldots+\sqrt[n]{1}}}}$$ (where I've added the $1$ at the beginning because it looks nicer that way and I wanted to). Then we have that $r_1=1$ is a root of the polynomial $p_1(x) = x-1$ and that (by recursion, if you want) $r_n$ is a root of $p_n(x) = p_{n-1}(x)^n-1$.
Unfortunately my Galois theory is a bit rusty, and I don't have time right now to review the subject, but I think it would be pretty straightforward to show that the polynomial induce a sequence of field extensions over $\mathbb{Q}$ which have strictly increasing degree, which should imply (I think) that the limit of the sequence of the $r_n$, if it exists, is transcendental.
• It is possible for a sequence $a_n$ of algebraic numbers of strictly increasing degree to converge to an algebraic number, or even a rational number. One example is $a_n:=2^{1/n}$. I have no reason to expect this behavior in this particular problem, however. – Julian Rosen Jul 31 '13 at 1:20