Calculating $2\sqrt{2}$ to Infinity

[maxkor]

Calculate $2\sqrt{2\sqrt[5]{2\sqrt[8]{2\sqrt[11]{2 \cdots}}}}$.
I know only that $...=2^{1+{1\over2}+{1\over10}+{1\over80}+{1\over880}+\ldots}$

 

[HOI]

In order to answer that we will need to know exactly how the exponent are calculated. You have roots of 1, 2, 5, 8, 11. But what next? I don't see any obvious pattern.

 

[maxkor]

14,17,...

 

[HOI]

Really? I saw that 2+ 3= 5, 5+ 3= 8, 8+ 3= 11 but 1 to 2 does not fit that.

 

[maxkor]

If you want:
calculate $2\ \cdot \ \sqrt{2\sqrt[5]{2\sqrt[8]{2\sqrt[11]{2 \cdots}}}}$.
2 * "that sqrts"

 

[topsquark]

So you need
\(\displaystyle 2 \sqrt{2} \sqrt{ \Pi _{n = 0}^{\infty} 2^{1/(5 + 3n)} }\)

I have no proof but Mathematica says this does not converge.

-Dan

 

[maxkor]

No,
\(\displaystyle 2 \sqrt{2} \sqrt{ \Pi _{n = 0}^{\infty} 2^{1/(5 + 3n)} } \neq 2^{1+\frac12+\frac1{2\cdot5}+\frac1{2\cdot5\cdot8}+\frac1{2\cdot5\cdot8\cdot11}+\cdots}\)

 

[topsquark]

Yes, I was reading that wrong.

Okay, I can't finish it but I can get it started.
\(\displaystyle P = 2\sqrt{2\sqrt[5]{2\sqrt[8]{2\sqrt[11]{2 \cdots}}}}\)

\(\displaystyle P = 2 \cdot 2^{1/2 + 1/2 \cdot 1/5 + 1/2 \cdot 1/5 \cdot 1/8 + 1/2 \cdot 1/5 \cdot 1/8 \cdot 1/11 + \text{ ...}}\)

\(\displaystyle ln(P) = ln(2) + \dfrac{1}{2} ln(2) + \dfrac{1}{2} \cdot \dfrac{1}{5} ln(2) + \dfrac{1}{2} \cdot \dfrac{1}{5} \cdot \dfrac{1}{8} ln(2) + \dfrac{1}{2} \cdot \dfrac{1}{5} \cdot \dfrac{1}{8} \cdot \dfrac{1}{11} ln(2) + \text{ ...}\)

\(\displaystyle ln(P) = ln(2) \left ( \dfrac{1}{2} + \dfrac{1}{2} \cdot \dfrac{1}{5} + \dfrac{1}{2} \cdot \dfrac{1}{5} \cdot \dfrac{1}{8} + \dfrac{1}{2} \cdot \dfrac{1}{5} \cdot \dfrac{1}{8} \cdot \dfrac{1}{11} + \text{ ...} \right )\)

So we have the form
\(\displaystyle ln(P) = ln(2) \left ( \sum_{n = 0}^{ \infty } a_n \right )\)
where \(\displaystyle a_n = \dfrac{ a_{n - 1} }{3n + 2}\), with \(\displaystyle a_0 = \dfrac{1}{2}\)

The trouble is that the series is neither arithmetic nor geometric. The defining recurrence is \(\displaystyle (3n + 2)a_n - a_{n - 1} = 0\) 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 \(\displaystyle a_n\) 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