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We can easily prove that the equation of variable $x$ $$(E_{n}): \frac{x(\ln x)^n}{1+x}=\frac{e}{2(e+1)}$$ has a unique solution $u_{n}$ in $[1,e]$ for all integers $n$ greater than $1$. Let's call it $u_{n}$.

Can you help me prove that

$$\lim_{n \to \infty} (\ln u_{n})^n=\frac{1}{2}$$


Note : This is a homework question. The general question which we aim to prove is that there exists a sequence $(\varepsilon_{n})_{n \geq 2} $ tending to $0$ and satisfying for all $n \geq 2$ : $u_{n}=e-\frac{e}{2n}+\frac{1}{n}\varepsilon_{n}$. I have been able to prove that $\varepsilon_{n}=n(u_{n}-e)+\frac{e}{2}$ and answering the question I am asking would allow me to prove that $\lim_{n \to \infty}n(u_{n}-e)=-\frac{e}{2}$ (since I know that $\lim_{n \to \infty}\frac{(\ln u_{n})^n-1}{n(u_{n}-e)}=\frac{1}{e}$ from a previous question). Hope it's not too unclear. Thank you.

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What you seek to prove is equivalent to the fact that $u_n\to\mathrm e$. Can you show that $u_n\gt v$ for $n$ large enough, for every $v\lt\mathrm e$? – Did Dec 25 '11 at 20:46
Well, this is something I have actually already proved ! But how can it help me ? – user20010 Dec 25 '11 at 22:26
up vote 1 down vote accepted

You said in comments that you were able to show that $u_n\to\mathrm e$. Since there exists a nonzero constant $c$ such that, for each $n$, equation $(E_n)$ reads as $(\log u_n)^n=c\cdot(1+u_n)/u_n$, the asymptotics $(\log u_n)^n\to c\cdot(1+\mathrm e)/\mathrm e$ follows.

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Thank you. How stupid I am ! – user20010 Dec 26 '11 at 18:22

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