17
$\begingroup$

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?

$\endgroup$
8
  • 2
    $\begingroup$ Great question! (+1) $\endgroup$ – user 1591719 Jul 30 '13 at 22:09
  • 1
    $\begingroup$ What closed form do you expect if the Inverter fails? $\endgroup$ – Hagen von Eitzen Jul 30 '13 at 22:18
  • 1
    $\begingroup$ @Chris'ssis Why? $\endgroup$ – Did Jul 30 '13 at 22:35
  • 1
    $\begingroup$ @Did If I answered the question I'd describe myself and I don't wanna do that, not here, not now. :-) $\endgroup$ – user 1591719 Jul 30 '13 at 22:51
  • 2
    $\begingroup$ @Chris'ssis ?? Whatever. $\endgroup$ – Did Jul 30 '13 at 22:53
4
$\begingroup$

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.

$\endgroup$
3
  • 2
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.