Set
$\bullet\text{ } a_1 = 2$
$\bullet \text{ }a_{j+1} = \frac{a_{j}}{\log(a_{j})}$
Is $(a_{j})_{j=1}^{\infty}$ bounded or is it unbounded?
If $a_{1}$ were to be $e$ then it would be obviously bounded.
If the above sequence were to be bounded then it is clear that $\exists \delta >0$ so that $\min_{j \in \mathbb{N}}|a_{j}-1| \geq \delta$
Update it seems to be converging to $e$.
It converges to $e$. Set $f(x) = \frac{x}{\log(x)}$ then it has a mimima at $x = e$ which is $e$. Thus $(a_{j})_{j=2}^{\infty} \geq e$.
We also have $a_{2} \geq a_{3} \geq...$ as in Brian's comment. Thus by M.C.T the above sequence converges to some $p \in [e, \frac{2}{\log(2)})$. If $p > e$ then note that if some d is sufficiently close to $p$ then $\frac{d}{\log(d)} < p$. Hence $p \leq e$ and therefore $p = e$.