What does this sequence converge of recursive logarithms to?

Question

Does anyone know what this sequence converges to?

$t_1 = \ln(n)$

$t_2 = \ln(n - t1) = \ln(n - \ln(n))$

$t_3 = \ln(n - t2) = \ln(n - \ln(n - \ln(n)))$

$...$

$t_k = \ln(n - t_{k-1})$

Notice that $t_1 < t_2$, $t_2 > t_3$, $t_3 < t_4$, ... and that $|t_k - t_{k+1}| \to 0$

Answer 1

Just as Gerhard S. commented, then you need to solve for $x$ the equation $e^x+x=n$, the solution of which being given in terms of Lambert function $$x=n-W(e^n)$$

Assuming $n$ to be "large", a very good approximation is given by $$x=\left(1-\frac{1}{n}\right) \log (n)$$ For $n=10$, the "exact" value would be $\approx 2.07058$ while the approximation gives $\approx 2.07233$.

Answer 2

In the paper : https://fr.scribd.com/doc/34977341/Sophomore-s-Dream-Function , page 13, §."Relationship with the Lambert W function", one can found an expansion in infinite series which corresponds to your sequence.

The notations $x$ in the paper and $n$ in your question are related as : $$n=\ln(x)$$

The condition of convergence given in the paper is $\quad x\geq e\quad$ thus, corresponding to $n\geq 1$.

Putting $x=e^n$ into the series given in the paper : $\quad\to\quad W(e^n)=n-t_\infty (n)$ $$t_\infty (n)=n-W(e^n)$$ This is consistent with Claude Leibovici's result.