TITLE: Convergence rate of $a_{n+1}=\ln(a_n+1)$
QUESTION [9 upvotes]: Let $a_0=1$ and $a_{n+1}=\ln(a_n+1)$. The goal is to find 
$$
(\star)\quad \lim_{n\to \infty}\frac{n(na_n-2)}{\ln n}
$$
It is easy to see that $a_n\to 0$. 
In addition, we can prove that 
$$
\lim_{n\to \infty}na_n=2.
$$
Indeed, one can obtain by  Stolz's theorem that 
\begin{align} \lim_{n\to\infty}\frac{n}{\frac{1}{a_n}} &= \lim_{n\to\infty}\frac{(n+1)-n}{\frac{1}{a_{n+1}}-\frac{1}{a_n}} \\ &=\lim_{n\to\infty} \frac{1}{\frac{1}{\ln(1+a_n)}-\frac{1}{a_n}} \\ &= \lim_{n\to\infty}\frac{a_n\ln(1+a_n)}{a_n-\ln(1+a_n)}  \\ &=2 , \end{align}
How to estimate the further limit $(\star)$?

REPLY [3 votes]: Let $L$ denote the limit in the OP. Then
$$
a_n=\frac{2}{n}+\frac{L\log n}{n^2}+\mathcal O(n^{-3}\log n)
$$
and
$$
\log(a_{n-1}+1)=\frac{2}{n}+\frac{L\log n}{n^2}+\frac{\frac23-L}{n^3}+\mathcal O(n^{-3}\log n)
$$
Insisting that both expansions are identical requires $L=\frac23$. A similar technique can be used to obtain higher order corrections.