Nested logarithm ratio limit

Question

Let $$f(x)=x+\ln(x+\ln(x+\cdots)))$$

Prove for $a\ge 1$, $$\lim_{n\to\infty} \frac{f(a^n)}{f(a^{n-1})}=a$$

I found this on my own so I have no idea if this is fairly well-known. So links to relevant answers or material is also appreciated.

Answer 1

Note that $f(x) = x + \ln(f(x))$. We in fact have: \begin{eqnarray} \ln(f(x)) - f(x) &=& -x\\ e^{-f(x)}f(x) &=& e^{-x}\\ e^{-f(x)}(-f(x)) &=& -e^{-x}\\ f(x) &=& -W(-e^{-x}) \end{eqnarray} where $W(z)$ is some branch of the Lambert W function. I believe what you are looking for is the branch $W_{-1}(z)$, which would make $f(x)$ real-valued for $x>0$ and diverge to infinity as $x$ goes to infinity.

To answer your particular question, we first observe: $$ \lim_{x\rightarrow\infty} \frac{f(x)}{x} =\lim_{x\rightarrow\infty} \frac{x+\ln f(x)}{x}= 1 + \lim_{x\rightarrow\infty} \frac{\ln f(x)}{x} = 1 + \lim_{x\rightarrow\infty} \frac{f'(x)}{f(x)} = 1 $$ by L'Hopital's rule. The last limit goes to 0 because $$ f'(x) = \frac{d}{dx} -W_{-1}(-e^{-x}) = \frac{W_{-1}(-e^{-x})}{W_{-1}(-e^{-x}) + 1} = \frac{f(x)}{f(x) - 1} $$ which clearly goes to 1 as $x$ approaches infinity, and hence $f'(x)/f(x) \rightarrow 0$.

Hence $f(x)$ is asymptotically equivalent to $x$, so $f(a^n) \sim a^n$ and your limit follows.