I'm dealing with this sequence. Let $a_0\in(1,\infty)$. Define $$a_{n+1}=1+\ln a_n$$
I have seen that $$\frac{a_{n+1}-1}{a_n-1}=\frac{\ln a_n}{a_n-1}$$ By differentiating $f(t)=\ln t-t+1$ I have seen that the ratio is bounded by a constant $<1$, so the sequence converges to $1$, at least as quick as a geometric progression.
Question: Is there a better asymptotic estimation for this sequence? I feel that it converges to $1$ much faster than a geometric progression to $0$.
Motivation: Given a natural $N$, let be $S(N)$ the sum of the digits of $N$. Let $$n=\min\{k\in\Bbb N:S^k(N)\text{ has one digit}\}$$ (where the exponent $k$ is for composition). I want to find a simple estimation for $n$.
$a_n$ behaves like $1+\frac2n$. Convergence is very slow...
Proof: Let $b_n=a_n-1$. Then the recursion is $b_{n+1}=\ln(1+b_n).$ Therefore the sequence $b_n$ is strictly decreasing with limit $0$. Consider now that $\ln(1+x)=x-\frac12x^2+o(x^2)$ as $x\to0.$ This gives $$\frac1{b_{n+1}}=\frac1{\ln(1+b_n)}=\frac1{b_n(1-\frac12b_n+o(b_n))}=\frac1{b_n}+\frac12+o(1).$$ This means that $\lim_{n\to\infty}\left(\frac1{b_{n+1}}-\frac1{b_n}\right)=\frac12$. By Cesaro's theorem we conclude that $$\lim_{n\to\infty}\frac1{n} \left(\frac1{b_n}-\frac1{b_0}\right)=\lim_{n\to\infty}\frac1{n} \left(\left(\frac1{b_n}-\frac1{b_{n-1}}\right)+ \cdots+\left(\frac1{b_1}-\frac1{b_{0}}\right)\right)=\frac12.$$Hence $b_n=\frac2n(1+o(1))$ and $a_n=1+\frac2n(1+o(1))$.