Find a value of
$$\lim_{n\rightarrow\infty}n\left ( e- e^{\frac{1}{e}}\uparrow\uparrow n \right )$$
For your information$,\quad\uparrow\uparrow$ is a tetration defined as
$$a\uparrow\uparrow n:=\underbrace{a^{a^{\cdot^{\cdot^{\cdot^{a}}}}}}_{n}$$
I think we must use the function $\ln x$ here then use the Laurent series of $\ln x.$ But I couldn't find an extension formula of $\ln e^{\frac{1}{e}}\uparrow\uparrow n,$ I need to the help, thanks a real lot !
Best Answer
To put that into a more reasonable (though less prone to enthusiastic upvotes) form: We have $x_1=e^{1/e}$ and $x_{n+1}=e^{x_n/e}$, and we're looking for $\lim_{n\to\infty}n\,(e-x_n)$. Letting $y_n=1-x_n/e$, some elementary algebra gives $$y_{n+1}=1-e^{-y_n}.$$ The precise value of $y_1$ is not so important, as long as it is positive. Then, $y_n$ is monotone decreasing, and positive, so it must have a limit $y$ satisfying $y=1-e^{-y}$, i.e. we must have $\lim_{n\to\infty}y_n=0$. Now some more elementary algebra gives $$\frac1{y_{n+1}}-\frac1{y_n}=\frac{e^{-y_n}+y_n-1}{y_n\,(1-e^{-y_n})}\to\frac12$$ as $n\to\infty$, and by the Stolz–Cesàro theorem, we have $$\lim_{n\to\infty}\frac1{n\,y_n}=\frac12.$$ Substituting the definition of $y_n$, we obtain $$\lim_{n\to\infty}n\,(e-x_n)=2\,e.$$ Convergence is rather poor, the error is $O\left(\frac{\log n}n\right)$.