Limit of a nested radical containing fractions

Question

How can I give a proof of the following inequality? $$\displaystyle L=1+\sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+\sqrt[4]{\frac{1}{4}}+...}}$$ $$L\le\sqrt[3]{4\pi}$$ Numerical computations suggest it's true, but is it possible to give an analytical evidence?

Answer 1

First note that the nested root $\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}$ converges to $\frac{1}{2}(1+\sqrt{5})$ (see (*) below).

To get the proposed inequality, we crudely estimate:

$$ \begin{align} L &= 1+\sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+\sqrt[4]{\frac{1}{4}+\cdots}}}\le1+\sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+\sqrt[4]{1+\sqrt[5]{1+\cdots}}}}\\ &\le1+\sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+\sqrt{1+\sqrt{1+\cdots}}}}=1+\sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+\frac{1+\sqrt{5}}{2}}}<2.323<\sqrt[3]{4\pi}. \end{align} $$

Better estimates can be obtained by introducing the $1$s a later position. I don't know where the expression $\sqrt[3]{4\pi}\approx 2.3249$ comes from but if we can produce better estimates as simple as that, it seems to me like that is not important anyway.

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Note:

(*) The value can be determined using the recurrence relation $x_{n+1}=\sqrt{1+x_n}$.