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While on chat, an interesting limit popped out: $$\sqrt[2]{1+\sqrt[3]{1+\sqrt[4]{1+\cdots}}}\approx 1.5176001678777188...\lt\phi$$ robjohn determined its value for fifty places, and Inverse Symbolic Calculator yields nothing.

Is there a closed form for this nice limit?

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    $\begingroup$ Great question! (+1) $\endgroup$ – user 1357113 Jul 30 '13 at 22:09
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    $\begingroup$ What closed form do you expect if the Inverter fails? $\endgroup$ – Hagen von Eitzen Jul 30 '13 at 22:18
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    $\begingroup$ @Chris'ssis Why? $\endgroup$ – Did Jul 30 '13 at 22:35
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    $\begingroup$ @Did If I answered the question I'd describe myself and I don't wanna do that, not here, not now. :-) $\endgroup$ – user 1357113 Jul 30 '13 at 22:51
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    $\begingroup$ @Chris'ssis ?? Whatever. $\endgroup$ – Did Jul 30 '13 at 22:53
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Let $$r_n := 1+\sqrt[2]{1+\sqrt[3]{1+\sqrt[4]{\ldots+\sqrt[n]{1}}}}$$ (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 $p_n(x) = p_{n-1}(x)^n-1$.

I suggest to study this sequence of polynomials instead of your complicated sequence of nested roots.

Edit: Removed the following wrong claim (after a nice counterexample by Pink Elephants):

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.

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    $\begingroup$ 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. $\endgroup$ – Julian Rosen Jul 31 '13 at 1:20
  • $\begingroup$ @PinkElephants Thanks for your comment, I will correct my answer. $\endgroup$ – Daniel Robert-Nicoud Jul 31 '13 at 7:46
  • $\begingroup$ @MatemáticosChibchas I am pretty sure my sequence of polynomials works, if that's what you're meaning. $\endgroup$ – Daniel Robert-Nicoud Jul 31 '13 at 7:47

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