Is this recurrence sequence in logarithmic form bounded? Consider the recurrence sequence: $a_{n+1}=\ln a_n$. Whatever the value of the first term is, the sequence go down very quickly, and the last term would be less than or equal to $0$. To continue the sequence, we can add absolute value sign into logarithm, then $a_{n+1}=\ln |a_n|$. Maybe for many values of the first term, the sequence get into chaos.
I think the sequence is unbounded because the terms of sequence probably dense near the point of $x=1$. So how to prove it?
Related question: If there is a number $α$, $\lim\limits_{n \to +\infty}\dfrac{a_n}{n^α}=0$, just like $\lim\limits_{n \to +\infty}\dfrac{\tan n}{n^8}=0$? When $n$ is enough large ($n>10^5$), $\dfrac1n\sum\limits_{i=1}^{n}a_i$ is approximately equal to $-0.569$, so $\dfrac1n\sum_{i=1}^{n}a_i$ converge？
 A: Ok, I'll assume you want to deal with real numbers here as it makes the analysis much easier. If you want some help with complex-valued sequences, I would recommend tagging your post with "complex-analysis". With this in mind, I'll examine the sequence $a_{n+1} =\ln( |a_{n} |)$ (Because the sequence $a_{n+1} =\ln( a_{n})$ will eventually generate complex values).
Will this sequence stay bounded? Probably not. If we start with some $a_0 > 1$, then after enough terms, we will end up with some value $a_n<1$. If this number is small enough,  then $|a_{n+1}|=|\ln(a_n)|\gg a_n$. Take for instance $a_n = 0.00001$. Then $|a_{n+1}|\approx 11.513$ which is roughly $10^6$ times larger than $a_n$. And for $a_n$ that are even smaller, $|a_{n+1}|$ will be even larger. Since this sequence is quite chaotic (see below) it stands to reason that after many terms we will hit very small values of $a_n$ that will correspond to large negative values for $a_{n+1}$. In short, there is no definite lower bound we can place on $a_n$, so it is unbounded. Then, because $a_n$ is unbounded, the sequence of partial sums $s_{n} =\sum ^{n}_{k=0} a_{k}$ definitely does not converge, but what about $$s_{n} =\sum ^{n}_{k=0}\frac{a_{k}}{k^{\alpha }}$$
I played around with it a little bit in Python and it appears you hit this magic point at $\alpha \approx 1.7$ where the sum seems to converge. I know this isn't a formal proof, but playing around with it a bit in Python or Mathematica will give you a good sense as to whether the sum will converge or not. 
As you indicated in your post, $a_n$ is quite chaotic. Shown below is a plot of $a_{n+1} =\ln( |a_{n} |)$ with $a_0=2$.
$a_n$">
Below is a plot of the partial sums, $s_n$, of $a_{n+1}=\frac{\ln(|a_n|)}{n^\alpha}$, with $\alpha=1.7$.

And the Python code I used to generate these graphs (if you want to play with it yourself):
import numpy as np
import matplotlib.pyplot as plt
plt.rcParams.update({'font.size': 22})
plt.rcParams["figure.figsize"] = (15,10)
plt.style.use('seaborn-bright')

a = [2] #this is the intial value, a0. 
N=1000 #the number of terms to compute
alpha=1.7 #The exponent alpha in the denominator of the partial sum formula. 
        #Set this equal to zero if you just want the original sequence.
for k in range(1,N):
    a.append(np.log(abs(a[k-1]))/k**alpha) #This computes the terms in the sequence of a
n = np.arange(0,N,1)
plt.scatter(n,a,marker='.') #this plots the sequence
plt.xlabel('$n$')
plt.ylabel('$a_n$')
plt.axhline(color='black')
plt.show()
s = []
for k in range(0,N):
    s.append(np.sum(a[0:k])) #this computes the kth partial sum
plt.scatter(n,s,marker='.')  #this plots the partial sums
plt.xlabel('$n$')
plt.ylabel('$s_n$')
plt.show()

A: This is not a complete answer, but I can prove that the behavior of this sequence is extremely dependent on what you define $a_0$ as. See also my answers here and here. I would note that it's a trivial modification of my Theorem 2 in the first linked answer that for any $A\in\mathbb{R}$, the set $\{a_0\in\mathbb{R} : \exists n\in\mathbb{N} \mbox{ such that } a_n=A\}$ is dense.

Let $a_n(x) = \log |a_n(x)|$, with initial value condition $a_0(x)=x$. Then for any $\alpha$, the following set is uncountable and dense in $\mathbb{R}$:$$
\{x : \frac{a_n(x)}{n^\alpha}\mbox{ unbounded}\}
$$
Futhermore, the set $\{x : a_n(x)\mbox{ bounded}\}$ is also uncountable and dense. To prove the first claim (as I noted above) it suffices to show that the set is uncountable. Let $S=\{s_1,s_2,\cdots\}\subset \mathbb{N}$ be any sequence such that $s_{n+1} \ge 2 s_n$ for all $n$. There is an uncountable number of such sequence, and for each one, we can construct $x$ such that $a_n(x)$ grows super-exponentially: Let $\epsilon_n =\begin{cases}1 & n\notin S\\ -1 & n\in S\end{cases}$ and define $$
x = e^{\epsilon_1 e^{\epsilon_2 e^{\cdots}}}
$$
Then we observe $$
a_n(x) = \epsilon_{n} e^{\epsilon_{n+1} e^{\epsilon_{n+2} e^\cdots}}
$$
If $\epsilon_{n+1},\dots,\epsilon_{n+k}$ are all $1$, then one can see $$
|a_n(x)| = e^{\epsilon_{n+1} e^{\cdots^{\epsilon_{n+k} e^{\epsilon_{n+k+1} e^{\cdots}}}}} \ge e^{\epsilon_{n+1} e^{\cdots^{\epsilon_{n+k}\cdot 0}}} = {^{k-1} e}
$$
where the left superscript represents tetration. In this case $\epsilon_{n}\ne 1$ if and only if $\epsilon_n =s_m$ for some $m$. Thus: $$
a_{s_m}(x) \ge {^{(s_{m+1} - s_m - 1)}e} \ge {^{s_m -1} e}
$$
For $s_m$ sufficiently large, we have ${^{s_m -1} e} > s_m^\alpha$, so therefore $a_n(x)/n^\alpha$ is unbounded. Indeed, we have that for any sequence $b_n$ growing slower than tetration, $a_n(x)/b_n$ is unbounded. Furthermore, one can use a similar construction (using faster growing $s_n$) to show that for any sequence $t_n$, the set $$
\{x : \frac{a_n(x)}{t_n}\mbox{ unbounded}\}
$$
is uncountable and dense.
One can similarly find an uncountable dense set such that $a_n$ is bounded. Let $s_n$ be an increasing sequence such that $s_{n+1} - s_n$ is bounded (say $\le M$) for all $n$. Again, there are uncountably many such sequences. Define $x$ similarly to above. If $\epsilon_{n+1},\dots,\epsilon_{n+k}$ are all $1$ and $\epsilon_{n+k+1} = -1$, then we have $$
|a_n(x)| = e^{\epsilon_{n+1} e^{\cdots^{\epsilon_{n+k} e^{\epsilon_{n+k+1} e^{\cdots}}}}} \le e^{\epsilon_{n+1} e^{\cdots^{\epsilon_{n+k} e^{\epsilon_{n+k+1} \cdot 0}}}} = {{^{k+1}}e}
$$
By assumption, there are no runs of $1$'s among the $\epsilon_n$'s of length greater than $M$, and hence $|a_n(x)|\le {^{M+2}e}$ for all $n$.
You can also show the set of cyclic points is dense (though this is countable), and even that there exists a particular value $x_0$ such that $\{a_n(x_0)\}$ is dense - I actually expect almost all starting values have this property but I haven't thought of a way to prove it yet.
I definitely expect there is some type of behavior that we can almost always expect for $a_n(x)$, but it's not totally clear to me what that is.

As for $\frac1n \sum_{k=1}^n a_n$, this is much harder. I don't know how to prove convergence except in cases where $a_n$ is cyclic or almost so.
