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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 !

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    $\begingroup$ is the $n$ part of what you are taking the limit of? $\endgroup$ Feb 26, 2021 at 3:16
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    $\begingroup$ Apart from that there is no convention whether $0$ is a natural number , what on earth has that to do with this question ? $\endgroup$
    – Peter
    Feb 26, 2021 at 9:13
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    $\begingroup$ Essentially, I understand this to mean: find $\lim_{n\to\infty}n(e-x_n)$, where $x_1=\alpha$, $x_{n+1}=\alpha^{x_n}$ and $\alpha=e^{1/e}$, is this correct? That limit would be simply $2e$. $\endgroup$
    – NoNames
    Feb 26, 2021 at 9:19
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    $\begingroup$ @Angelo please do not change '\lim to \lim\limits in titles. Firstly you are disagreeing with the OP who has repeatedly rolled back your edit. Secondly Displaystyle mathematics uses up more vertical space than is needed in a list of questions. Thirdly its probably better if you don't do it inside the body, when it appears in between a pair of single dollar signs $...$ $\endgroup$ Mar 4, 2021 at 7:31
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    $\begingroup$ The only reason I have not reverted your edit is because it is wrong to continuously bump a post for a minor edit (which is precisely what you have been doing). Please stop making titles use more vertical space than is needed. Regarding using it in the body, It makes line spacing bad. $\endgroup$ Mar 5, 2021 at 8:54

1 Answer 1

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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 $O\left(\frac{\log n}n\right)$.

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    $\begingroup$ How have you obtained that the error is $O\left(\frac{\log n}n\right)$? Please, could you tell me it? $\endgroup$
    – Angelo
    Mar 9, 2021 at 6:12
  • $\begingroup$ Nonames, please, could you tell me why the error is $O\left(\frac{\log n}n\right)$ ? $\endgroup$
    – Angelo
    Mar 15, 2021 at 6:09

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