2
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$$ \sqrt[2]{2\sqrt[3]{2\sqrt[5]{2\sqrt[7]{2\sqrt[11]{2\sqrt[13]{2\sqrt[17]{2...\sqrt[p]{2... \infty }}}}}}}} $$

Where $p$ is prime.

General term can be written as:

$$ \prod\limits_{i=1}^n 2^{\frac{1}{p_1*p_2...p_n}} $$

How to find its value or at least an asymptotic upper bound ?


Please feel free to add suitable tags.

  • 2
    it's between $\sqrt 2$ and $2$ (when is an upper bound "asymptotic" ??)2017-02-22
  • 1
    About $1.6304046418509$.2017-02-22

1 Answers 1

4

When taking the base-$2$ logarithm, you get a sum that converges faster than the series for $e$.

The first iterates are

$$0.5 \\ 0.666666666667 \\ 0.7 \\ 0.704761904762 \\ 0.705194805195 \\ 0.705228105228 \\ 0.705230064054 \\ 0.70523016715 \\ 0.705230171632 \\ 0.705230171787 \\ 0.705230171792$$

When you truncate the sum, the sum of the tail will not exceed twice the first omitted term*. Then

$$S_n>S_\infty-\frac2{p_1\cdot p_2\cdots p_{n}}.$$

From the asymptotics of the primorial,

$$S_\infty-S_n=\frac{e}{n^{n(1+o(1))}}.$$


*A better approximation of this factor $2$ is $1+S_\infty$, or better (and much closer to $1$), $1+\dfrac2{p_{n+1}}$.

  • 0
    The [primorial](http://oeis.org/A064648) gave a good insight to it, thanks a lot :)2017-02-22