I am solving this question:
Let $x_0=1$ and $$x_{n+1}=\log\left(\dfrac{e^{x_n}-1}{x_n}\right)$$ Compute the limit :$$\lim_{n\rightarrow\infty}2^nx_n$$ The answer is NOT 1 because
it is easy to find that $$\dfrac{x_{n+1}}{x_{n}}=\dfrac{1}{x_{n}}\log\left(\dfrac{e^{x_n}-1}{x_n}\right)>\dfrac{1}{2}$$ because $x_{n}\downarrow 0$. So $2^{n+1}x_{n+1}>2^nx_{n}>1$ as $n\rightarrow\infty$
Using methods of complex analysis, I found the answer equals $f(1)$ where $f(x)$ is analytic function with $f(0)=0$ and satisfies the equation:$$f\left(\log\left(\dfrac{e^x-1}{x}\right)\right)=\dfrac{1}{2}f(x)$$ But I met difficulty solving this equation. My questions are:
- Is there a closed form of the solution?If not, what is its numerical solution?
- Is there another method to estimate the asymptotic behavior of $x_n$?