6
$\begingroup$

Context

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

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

then

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

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

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

then

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


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.

$\endgroup$
  • $\begingroup$ 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
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.