An asymptotic expansion for $u_{n+1} = u_n^2 + u_n$

Question

I'm completely stuck finding an equivalent of the following recurring sequence.

$\left\{\begin{matrix} u_0 > 0\\ u_{n+1} = u_n^2 + u_n \end{matrix}\right.$

The problem suggests to use the sequence $x_n = 2^{-n} \ln(u_n)$ and $v_n = x_{n+1} - x_n$

I showed that

1. $u_n \to \infty$
2. $\sum v_n$ is convergent and so $x_n$ converges

Have you an idea in order to conclude and find an asymptotic expansion for $u_n$?

Answer 1

If $u_0=1$, the first eight terms are listed in sequence $A007018$ of $OEIS$.

Computing the first $25$, it seems that $$\color{blue}{\log(\log(u_n)) \sim (n-1) \, \log(2)}\tag 1$$ could be decent approximation.

This seems to be "normal" since the numbers are so large that, if it was $$u_{n+1}=u_n^2 \qquad \text{with} \qquad u_8=113423713055421844361000442$$ it would be $$\log(u_n)=\frac{\log (u_8)}{256}\, 2^n \quad \implies\qquad \log(\log(u_n))=n \log(2) -1.45095$$

For $n=25$, $(1)$ gives $16.6355$ while the exact value is $16.5709$.

Answer 2

From the fact that $$\lim_{n\to\infty} 2^{-n}\ln(u_n) = C$$ exists, you can conclude $$\ln(u_n) \sim C \cdot2^n.$$ Based on how the problem is posed, I would tend to assume that's all that is wanted.