Find a sequence such that this tower of of exponent is convergent 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.

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.
