I am looking at a sequence where $a_1>0$ and $$ a_{n+1}=a_n^{a_n}.$$ I want to figure out for what values of $a_1$ does this sequence converge. My guess is that it converges of $a_1\leq 1$ and diverges for $a_1>1$. I was able to show convergence for $a_1\leq 1$ since if $a_1=1$, then it is trivial, otherwise, we can make a change of variables $b_n=\ln(a_n)$ and show that $b_n$ converges, which implies $a_n$ converges. Then we have $$b_{n+1}=b_ne^{b_n}$$ Now if $b_1<0$, then $b_{n}<0$ for all $n$ since $b_{k+1}=b_ne^{b_k}<0$ as $e^{b_k}>0$ and $b_k<0$ inductively. However, we also have $b_{k+1}>b_k$ as $$b_{k+1}=b_ke^{b_k}>b_k$$ as we assume $b_k<0$ which implies $e^{b_k}\in(0, 1)$, and since $b_k<0$, this will lead to a larger number, so $b_k$ converges due to monotonic sequence theorem.
Now I am stuck on showing divergence. I know there are divergent values (for example $a_1=2$), but I have no clue about the general behavior about $a_1>1$, like values such as $a_1=1.0001$. I was wondering if I could have a hint. I was able to show that $b_{n+1}>b_n$ for all $n$, but I also want to show that there exists $N$ such that $b_n>M$ for $n>N$ which implies divergence. How do I do that? One possible approach for me is noticing that $$b_{n+1}=b_ne^{b_n}=b_{n-1}e^{b_n+b_{n-1}}=\cdots = b_0e^{\sum_{k=0}^n b_k}.$$ Now since $b_{k+1}>b_k$ for all $k$ and $b_1>0$, the sum clearly does not converge by the nth term test, so there is an $N$ such that $\sum_{k=0}^N b_k>\ln(M/b_0)$ for $n>N$. Then we have $$b_{n+1}>b_0e^{\ln(M/b_0)}=M$$ and since $b_n$ increases this shows $b_n$ diverges and thus $a_n$ diverges. Does this approach work?
