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.