What is the maximum value of $\ln(x+\ln(x+\ln(x+\cdots)))-\ln x$? This question is somewhat similar to my last set of infinite nests (see here) but this time I would like to attain an upper bound instead of the area, as $\int_1^\infty\ln x\,dx$ does not converge.

Consider $f_1(x)=\ln x$ and $f_2(x)=\ln(x+\ln x)$ such that $f_n$ satisfies the relation $$f_n(x)=\ln(x+f_{n-1}(x)).$$ Is there a minimum real number $k$ such that $f_\infty(x)\le\ln x+k$ and if so, is there a closed form?

Attempt
Let us first consider the equality. $$f_\infty(x)=\ln x+k\implies\ln(x+\ln x+k)=\ln x+k\implies x+\ln x=xe^k-k$$ From W|A, we get $k=-W(-e^{-x})-(x+\ln x)$. Its graph against $x$ can be found here.
Let's now try a numerical approach. I used Desmos for this - an example is here. \begin{array}{c|c}n&1&2&3&4&5&6&7&8&9&10\\\hline \text{(approx.) } k&0&0.313&0.404&0.437&0.451&0.456&0.458&0.458&0.459&0.459\end{array} This shows that $k\approx0.459$, hence there is strong evidence that such a number $k$ does exist for $f_{\infty}$.
Any ideas for the closed form?

After some successful discussions with @Holo here, we have derived a closed form for $k$.
The expression for $k(x)$ in the original post is incorrect in that we must take the negative branch of the Lambert $W$ function on $(-1,0)$; that is, $$k(x)=-W_{-1}(-e^{-x})-x-\ln x\tag{1}.$$ with $W_{-1}(-e^{-x})$ being the negative branch on $(1,\infty)$.
We want to find the maximum value of $f_{\infty}(x)-\ln x$. Differentiating $(1)$, we get $$k'(x)=-\frac{W_{-1}(-e^{-x})}{-e^{-x}(1+W_{-1}(-e^{-x}))}\cdot e^{-x}-1-\frac1x=-\frac1{1+W_{-1}(-e^{-x})}-\frac1x=0$$ for critical points. Therefore, we have $$x+1+W_{-1}(-e^{-x})=0\tag{2}$$ provided that $x\ne0$ and $W_{-1}(-e^{-x})\ne-1\implies x\ne1$ which is not a problem. 
From $(2)$, we finally arrive at the value of $x=e^{1-0}+0-1=e-1$, hence $$k=-W_{-1}(-e^{e-1})-e-1-\ln(e-1)=-(-e)-(e-1)-\ln(e-1)=\boxed{1-\ln(e-1)}$$
