# Find a limit of $\ln(u_n)/2^n$

Let $$u_0 \in \mathbb{N}^{*}$$ and $$(u_n)$$ such that $$u_{n+1}=1+u_n^2$$.

I've shown that the sequence $$\displaystyle \left(\frac{\ln\left(u_n\right)}{2^n}\right)$$ converges and I wonder if it possible to find its limit $$\ell$$

For example, if $$u_0=1$$ then $$\ell \approx 0.40735$$, if $$u_0=15$$ then $$\ell \approx 2.7103$$.

• A very good question (+1). Sep 12, 2019 at 22:02
• $2*\ln($ A076949 $)$ is what it converges to for $u_0 = 1$. In the OEIS, there is no closed form for this. Sep 12, 2019 at 23:05

With the $$2^n$$ in denominator, the limit is obtained quite fast.
Let $$u_0=10^k$$ and computing we get $$\left( \begin{array}{cc} k & \ell_k \\ 0 & 0.407354523 \\ 1 & 2.307584766 \\ 2 & 4.605220186 \\ 3 & 6.907755779 \\ 4 & 9.210340377 \\ 5 & 11.51292547 \\ 6 & 13.81551056 \\ 7 & 16.11809565 \\ 8 & 18.42068074 \\ 9 & 20.72326584 \end{array} \right)$$ and, as you can see, $$\ell_k \sim k \log(10)$$