Yes, I was reading that wrong.
Okay, I can't finish it but I can get it started.
[math]P = 2\sqrt{2\sqrt[5]{2\sqrt[8]{2\sqrt[11]{2 \cdots}}}}[/math]
[math]P = 2 \cdot 2^{1/2 + 1/2 \cdot 1/5 + 1/2 \cdot 1/5 \cdot 1/8 + 1/2 \cdot 1/5 \cdot 1/8 \cdot 1/11 + \text{ ...}}[/math]
[math]ln(P) = ln(2) + \dfrac{1}{2} ln(2) + \dfrac{1}{2} \cdot \dfrac{1}{5} ln(2) + \dfrac{1}{2} \cdot \dfrac{1}{5} \cdot \dfrac{1}{8} ln(2) + \dfrac{1}{2} \cdot \dfrac{1}{5} \cdot \dfrac{1}{8} \cdot \dfrac{1}{11} ln(2) + \text{ ...}[/math]
[math]ln(P) = ln(2) \left ( \dfrac{1}{2} + \dfrac{1}{2} \cdot \dfrac{1}{5} + \dfrac{1}{2} \cdot \dfrac{1}{5} \cdot \dfrac{1}{8} + \dfrac{1}{2} \cdot \dfrac{1}{5} \cdot \dfrac{1}{8} \cdot \dfrac{1}{11} + \text{ ...} \right )[/math]
So we have the form
[math]ln(P) = ln(2) \left ( \sum_{n = 0}^{ \infty } a_n \right )[/math]
where [math]a_n = \dfrac{ a_{n - 1} }{3n + 2}[/math], with [math]a_0 = \dfrac{1}{2}[/math]
The trouble is that the series is neither arithmetic nor geometric. The defining recurrence is [math](3n + 2)a_n - a_{n - 1} = 0[/math] and I have never solved one of these before. (I don't even know how to get Mathematica to solve it.) And the, once you get [math]a_n[/math] you still have to sum it.
This is a highly non-trivial problem for a specific value of n. There may be a way to make it simpler in the limit as n goes to infinity.
-Dan
