Find $ ? = \sqrt[3] {1 + \sqrt[3] {1 + 2 \sqrt[3] {1 + 3 \sqrt[3] \cdots}}} $

Solution 1:

Generalizing Ramanujan's Radical, you can get trivial results like this,

$$2=\sqrt[3]{4+1^2\sqrt[3]{10+3^{2}\sqrt[3]{16+5^{2}\sqrt[3]{22+7^{2}\sqrt[3]{28+9^{2}\sqrt[3]{...}}}}}}$$ $$3=\sqrt[3]{7+2^{2}\sqrt[3]{13+4^{2}\sqrt[3]{19+6^{2}\sqrt[3]{25+8^{2}\sqrt[3]{...}}}}}$$ Where the terms, $4,10,16,..$ are in an arithmetic progression with $d=6$. And so are the terms $7,13,19..$

While your stated question looks similar to Ramanujan's, it isn't. A closed form expression is highly unlikely.

Also, the above expressions are from;

$$n+1=\sqrt[3]{1+3n+n^{2}\sqrt[3]{1+3\left(n+2\right)+\left(n+2\right)^{2}\sqrt[3]{1+3\left(n+4\right)+\left(n+4\right)^{2}\sqrt[3]{...}}}}$$

Some approximations for the constant, one of them include;

$$\approx \sqrt{\frac{\sqrt{15}+\sqrt[3]{2}}{\sqrt{7}}}$$

At 6 decimal places, other approximations include; $12^{\frac{2}{15}},\sqrt{\frac{97}{50}},\exp\left(\frac{1}{3}-\frac{1}{500}\right),\sinh^{-1}\left(\frac{17}{9}+.0001\right),\frac{39}{28}$