The Euclid numbers are defined recursively by $e_1 = 2$, and $e_{n + 1} = \prod_{k = 1}^n e_k + 1$ for all $n \geq 1$. They also satisfy $e_{n + 1} = e_n^2 - e_n + 1$ and $e_{n + 1} - \frac{1}{2} = (e_n - \frac{1}{2})^2 + \frac{1}{4}$ for $n \geq 1$.
The sequence $a_n = 2^{-n} \log(e_n - 1/2)$ converges. To see this, we may observe (through some messy calculations) that $0 < \Delta a_n < \log(10/9) 2^{-(n + 1)}$. This implies that $\{a_n\}$ is strictly increasing and, by summing over $n$, that $\{a_n\}$ is bounded. Specifically, $a_{n + 1} < a_1 + \log(10/9)\sum_{k = 1}^\infty 2^{-(k + 1)} = \frac{1}{2} \log(5/3)$, so $\lim_{n \to \infty} a_n \leq \frac{1}{2} \log(5/3)$. Numerical evidence suggests that this is a reasonably good bound.
This can be generalized. The sequence $a_n = d^{-n} \log(e_n - 1/2)$ converges for all real $d \geq 2$. This can be seen by computing its forward difference: \begin{equation} \Delta a_n = d^{-(n + 1)} [\log(1 + 1/4(e_n - 1/2)^2) - (d - 2) \log(e_n - 1/2)]. \end{equation} If $d > 2$, then we can show that the sequence is eventually strictly decreasing. ($\Delta a_n = d^{-(n + 1)} D_n$ for suitably defined $D_n$, and it can be checked that $e^{D_n}$ will eventually be less than $1$; hence $D_n < 0$ eventually.) Since it is also positive it must converge. This tells us nothing about what it converges to, though.
Can we find "good" bounds of $\lim_{n \to \infty} d^{-n} \log(e_n - 1/2)$ for arbitrary reals $d \geq 2$?
In particular, I suspect that $\lim_{d \to \infty} \lim_{n \to \infty} d^{-n} \log(e_n - 1/2) = 0$. A bound that allowed us to prove this would be nice.
That's easy, it's just a product of limits. For $d>2$, you have $d^{-n} \log(e_n - 1/2)=(\frac{2}{d})^n 2^{-n} \log(e_n - 1/2)$. But $(\frac{2}{d})^n$ converges to zero while $2^{-n} \log(e_n - 1/2)$ converges to a finite limit. So $d^{-n} \log(e_n - 1/2)$ converges to zero.