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let equation $x^n+x=1$ have positive $a_{n}$.

show that $$\displaystyle\lim_{n\to \infty}\dfrac{n}{\ln{n}}(1-a_{n})=1$$

yesteday, I have post this and prove following $$\displaystyle\lim_{n\to \infty}a_{n}=1$$ How prove this limit $\displaystyle\lim_{n\to \infty}a_{n}=1$

Now I found This beautiful limtit. Thank you everyone can prove it

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Let $a_n=1-\varepsilon$. Then

$$(1-\varepsilon)^n - \varepsilon=0\implies n = \frac{\log{\varepsilon}}{\log{(1-\varepsilon)}} \sim \frac{1}{\varepsilon}\log{\frac{1}{\varepsilon}} \quad (\varepsilon \to 0)$$

One may show that, in this case, because this limit is equivalent to $n \to \infty$,

$$\varepsilon \sim \frac{\log{n}}{n} \quad (n \to \infty)$$

To see this, note that

$$\frac{1}{\varepsilon}\log{\frac{1}{\varepsilon}}\sim \frac{n}{\log{n}} \log{\frac{n}{\log{n}}} = \frac{n}{\log{n}} (\log{n}-\log{\log{n}})\sim n $$

Therefore,

$$\lim_{n \to \infty} \frac{n}{\log{n}} (1-a_n) = \lim_{n \to \infty} \frac{n}{\log{n}} \varepsilon = 1$$

as was to be shown.

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    $\begingroup$ Thank you,But $n\sim\frac{1}{\varepsilon}\log{\frac{1}{\varepsilon}}$? How can we prove must $\varepsilon \sim \frac{\log{n}}{n} \quad (n \to \infty)$? $\endgroup$ – math110 Sep 6 '13 at 1:52
  • $\begingroup$ @math110: this represents a bit of inverse thinking. The conditions $\varepsilon\to 0$ and $n \to \infty$ are equivalent, so we can "solve" for $n$ as I did in the first line above. From there, I showed how $\varepsilon$ behave in the third line. $\endgroup$ – Ron Gordon Sep 6 '13 at 2:08
  • $\begingroup$ Why not write $a_n=1-\varepsilon_n$? $\endgroup$ – Pedro Tamaroff Sep 6 '13 at 2:20
  • $\begingroup$ @PeterTamaroff: got seduced by the notation in the previous problem. No real good reason. $\endgroup$ – Ron Gordon Sep 6 '13 at 2:22
  • $\begingroup$ @RonGordon Ah, OK. At first, I thought the argument was another, so maybe it might improve clarity. $\endgroup$ – Pedro Tamaroff Sep 6 '13 at 2:24

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