$a_1>0$ and $a_{n+1}=\ln{(a_n+1)}$. Where is the largest term of the sequence $na_n$?

Question

Given $a_1>0$ and $a_{n+1}=\ln{(a_n+1)}$, the sequence $na_n$ is quite interesting.

It turns out that the sequence $na_n$ approaches $2$ (proof), and that $\begin{align}\lim_{n\to\infty}\frac{n}{\ln{n}}(na_n-2)\end{align}=2/3$ (proof).

So $na_n$ has some maximum value greater than $2$.

Let $f(x)=$ { $n$-value of $(na_n)_\text{max}$ when $a_1=x$ }.

An Excel simulation yields:

$f(3)=1$
$f(2)=3$
$f(1.5)=11$
$f(1)=118$
$f(0.8)=652$
$f(0.6)=10424$
$f(0.55)=28113$

What is a good approximation for $f(x)$ for small $x$ ?

I took the log of $f(x)$ and then used Excel's trendline tool to come up with $f(x)\approx ab^{1/x}$ where $a\approx 0.189$ and $b\approx 700$, but I doubt this is valid for very small values of $a_1$.

I'm wondering if there is a closed form approximation for $f(x)$.

Answer 1

By the results of this paper, $$ a_{n + 1} = \frac{2}{n}\left[ {1 + \frac{{\frac{1}{3}\log n - C(x)}}{n} + \mathcal{O}\!\left( {\frac{{\log ^2 n}}{{n^2 }}} \right)} \right] $$ as $n\to+\infty$, with $$ C(x) = \frac{2}{x} + \frac{1}{3}\log \left( {\frac{2}{x}} \right) + \frac{1}{{36}}x + \mathcal{O}(x^2 ) $$ as $x\to 0^+$. From this, we obtain $$ na_n - 2 = \frac{{\frac{2}{3}\log n - 2C(x) + 2}}{n} + \mathcal{O}\!\left( {\frac{{\log ^2 n}}{{n^2 }}} \right). $$ Maximizing the first term on the right-hand side yields $$ f(x) \approx \exp (3C(x) - 2) \approx \frac{{2{\rm e}^{ - 2} }}{x}{\rm e}^{6/x} \!\left( {1 + \frac{1}{{12}}x} \right) $$ as $x\to 0^+$.