• We already know that if we take a sequence $(x_n)\in{\mathbb R_+^*}^{\mathbb N}$ such that

$$x_n=O\left(\frac 1{n^2}\right)$$


$$\sum_{n=0}^\infty x_n <+\infty.$$

  • We also now that if we take for instance for all $n\in \mathbb N^*$

$$x_n=1+\frac 1{n^2}$$


$$\prod_{n=0}^\infty x_n <+\infty.$$

Can we find a sequence $(x_n)\in(1,\infty)^{\mathbb N}$ such that

$${{x_0}^{{x_1}^{{x_2}^{x_3}}}}^{\dots} = {{x_0}^{\left({x_1}^{\left({x_2}^{x_3^\cdots}\right)}\right)}}$$

is convergent?

I think the answer is yes if we take $(x_n)$ such that $\lim_{n\to\infty} x_n=1$ and $(x_n)$ converges to $1$ really fast. But I don't know how to exhibit such a sequence.

  • $\begingroup$ With $(x_n)\in(1,\infty)^{\mathbb N}$ do you mean that it must take integer values greater then one? $\endgroup$ – Jan Dec 22 '16 at 18:53
  • $\begingroup$ @Jan Yes, that is what I meant. $\endgroup$ – E. Joseph Dec 22 '16 at 20:57

The lame answer is to have $x_0=1$ or $x_0=0,x_1>0$. Then the result is trivial.

If $x_n=x_0$ for all $n$ and $x_0>0$, then it converges iff $e^{-e}\le x_0\le e^{e^{-1}}$, which actually does not require $\lim_{n\to\infty}x_n=1$. These are found in the Wikipedia for tetration. More information on the exact nature of convergence for $x_n\in\mathbb C$ may be found here and here and here.

From here, you can show that for any sequence with $e^{-e}\le x_0\le e^{e^{-1}}$ that is monotonically approaching $1$ will converge.

  • $\begingroup$ You are obviously right, and thank you for your answer. It allows me to specify what I needed (I edited). $\endgroup$ – E. Joseph Dec 22 '16 at 18:01
  • 1
    $\begingroup$ @E.Joseph Sure sure, go right ahead and challenge me :D $\endgroup$ – Simply Beautiful Art Dec 22 '16 at 18:11

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