How to prove $\lim_{n\to \infty}\frac{n}{\ln{n}}(1-a_{n})=1$?

Question

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

score 4 · Answer 1 · answered Sep 06 '13 at 01:45

4

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.

answered Sep 06 '13 at 01:45

Ron Gordon

138,521

1

Thank you,But $n\sim\frac{1}{\varepsilon}\log{\frac{1}{\varepsilon}}$? How can we prove must $\varepsilon \sim \frac{\log{n}}{n} \quad (n \to \infty)$? – math110 Sep 06 '13 at 01:52
@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. – Ron Gordon Sep 06 '13 at 02:08
Why not write $a_n=1-\varepsilon_n$? – Pedro Sep 06 '13 at 02:20
@PeterTamaroff: got seduced by the notation in the previous problem. No real good reason. – Ron Gordon Sep 06 '13 at 02:22
@RonGordon Ah, OK. At first, I thought the argument was another, so maybe it might improve clarity. – Pedro Sep 06 '13 at 02:24
I guess must $$g(n)<a_{n}<\dfrac{n}{n+1}$$?How find the simple form of $g(n)$.such $$\displaystyle\lim_{n\to\infty}g(n)=1$$ – math110 Sep 06 '13 at 02:27
@math110: I do not understand your question. – Ron Gordon Sep 06 '13 at 02:31
@RonGordon,I have edit,Now understand my question? – math110 Sep 06 '13 at 02:32
your mean? $$(1-e^{-n^2})^n-e^{-n^2}<0?$$ – math110 Sep 06 '13 at 02:46
@math110: no, I think I may have misunderstood. Why don't you ask this as a separate question? – Ron Gordon Sep 06 '13 at 02:51
I'm thinking about a simple method and beautiful methods to solve $$\lim_{n\to\infty}a_{n}=1$$ – math110 Sep 06 '13 at 02:57
@math110: The above asymptotic expression is a good start! – Ron Gordon Sep 06 '13 at 03:12

How to prove $\lim_{n\to \infty}\frac{n}{\ln{n}}(1-a_{n})=1$?

1 Answers1

Linked