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Assume $a\in\mathbb R,\,e^{-e}<a<e^{1/e}$ and $n\in\mathbb N$.

Let ${^n a}$ denote tetration: $${^0a}=1,\quad{^{(n+1)}a}=a^{\left({^n a}\right)}.\tag1$$ It is well known that under these assumptions, the following limit exists: $${^\infty a} = \lim_{n\to\infty}{^n a} = \frac{W\!\left(-\ln a\right)}{-\ln a} = e^{-W\left(-\ln a\right)},\tag2$$ where $W(z)$ is the Lambert $W$-function.

It is also known that $$\lim_{n\to\infty}\frac{{^\infty a} - {^{(n+1)} a}}{{^\infty a} - {^n a}} = {^\infty a} \cdot \ln a.\tag3$$

How can we prove that the following also holds?

$$\lim_{n\to\infty}\left(\frac 1 {{^\infty a} - {^n a}}-\frac{{^\infty a} - {^{(n+1)} a}}{{^\infty a} \cdot \left({^\infty a} - {^n a}\right)^2 \cdot \ln a}\right)=\frac{\ln a}2\tag4$$

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Following Nemo's idea, set $x=^\infty a-^na$, so that $x\rightarrow 0$ as $n\rightarrow\infty$. Then $$\frac{1}{^\infty a-^na}-\frac{^\infty a-^{(n+1)}a}{^\infty a(^\infty a-^na)^2\ln a}=\frac{1}{x}-\frac{1-\frac{^{(n+1)}a}{^\infty a}}{(^\infty a-^na)^2\ln a}=\frac{1}{x}-\frac{1-a^{^na-^\infty a}}{x^2\ln a}=\frac{1}{x}-\frac{1-a^x}{x^2\ln a}=\frac{a^x-x\ln a-1}{x^2\ln a}.$$

Expanding $a^x$ around $0$ as a power series $1+x\ln a+\frac{1}{2}x^2(\ln a)^2+O(x^3)$ gives that this expression, as $x\rightarrow 0$, is $$\frac{\frac{1}{2}x^2(\ln a)^2+O(x^3)}{x^2\ln a}=\frac{\ln a}{2}+O(x)\rightarrow\frac{\ln a}{2}.$$

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