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

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

Show that $$\lim_{n\to \infty}\frac{n}{\ln{(\ln{n}})}\left(1-a_{n}-\dfrac{n}{\ln{n}}\right)=-1$$

some hours ago,it prove that $$\lim_{n\to \infty}\frac{n}{\ln{n}}(1-a_{n})=1$$

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

this problem is from china paper(1993):http://www.cnki.com.cn/Article/CJFDTotal-YEKJ199301013.htm

and this paper poof is very ugly,so I want see other nice methods,I kown these can use Incremental estimation analysis,Thank you

Let $a_n=1-\frac1n(\log n)b_n$, then $(a_n)^n=1-a_n$ reads $n\log a_n=\log(1-a_n)$, that is, $$ n\log\left(1-\frac1n(\log n)b_n\right)=\log\left(\frac1n(\log n)b_n\right). $$ Since $a_n\to1$ and $\log(1-x)=-x+O(x^2)$ when $x\to0$, $$ \log\left(1-\frac1n(\log n)b_n\right)=-\frac1n(\log n)b_n+o\left(\frac1n\right). $$ Since $b_n\to1$, $\log b_n=o(1)$ hence $$ -(\log n)b_n+o(1)=\log\log n-\log n+o(1) $$ that is, $$ b_n=1-\frac{\log\log n}{\log n}+o\left(\frac1{\log n}\right). $$ This means that $$ 1-a_n-\frac{\log n}n+\frac{\log\log n}n=o\left(\frac1n\right), $$ which is equivalent to $$ \frac{n}{\log\log n}\left(1-a_n-\frac{\log n}n\right)=-1+o\left(\frac1{\log\log n}\right). $$ In particular, the following weaker conclusion holds: $$ \lim_{n\to\infty}\frac{n}{\log\log n}\left(1-a_n-\frac{\log n}n\right)=-1. $$

If we set $x=e^{-t}$, we get $$ x^n+x=1\iff -nt=\log\left(1-e^{-t}\right) $$ $$ \begin{align} -nt &=\log\left(1-e^{-t}\right)\\ &=\log(t)+\log\left(\frac{1-e^{-t}}{t}\right)\\ &=\log(t)-\frac{t}{2}+\log\left(\frac{\sinh(t/2)}{t/2}\right)\\ \frac12-n &=\frac{\log(t)}{t}+\frac1t\log\left(\frac{\sinh(t/2)}{t/2}\right)\tag{1} \end{align} $$ where, for $t\gt0$, $\frac1t\log\left(\frac{\sinh(t/2)}{t/2}\right)\le\min\left(\frac12,\frac{t}{24}\right)$.

On $(0,1)$ $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}t}\frac{\log(t)}{t} &=\frac{1-\log(t)}{t^2}\\ &\gt0\tag{2} \end{align} $$ therefore, $\frac{\log(t)}{t}$ monotonically increases from $-\infty$ to $0$ on $(0,1)$.

For $n\ge4$ $$ \begin{align} \frac{\log(\log(n)/n)}{\log(n)/n} &=\frac{\log(\log(n))-\log(n)}{\log(n)/n}\\ &=n\frac{\log(\log(n))}{\log(n)}-n\\ &\gt\frac12-n\tag{3} \end{align} $$ For $n\ge3$ $$ \begin{align} \frac{\log\left(\frac{\log(n)}{n}\left(1-\frac{\log(\log(n))}{\log(n)}\right)\right)}{\frac{\log(n)}{n}\left(1-\frac{\log(\log(n))}{\log(n)}\right)} &=\frac{\log(\log(n))-\log(n)+\log\left(1-\frac{\log(\log(n))}{\log(n)}\right)}{\frac{\log(n)}{n}\left(1-\frac{\log(\log(n))}{\log(n)}\right)}\\ &=-n+\frac{\log\left(1-\frac{\log(\log(n))}{\log(n)}\right)}{\frac{\log(n)}{n}\left(1-\frac{\log(\log(n))}{\log(n)}\right)}\\ &=-n-\frac{n\log(\log(n))}{\log(n)^2}\varphi\left(\frac{\log(\log(n))}{\log(n)}\right)\\ &\lt\frac12-n\tag{4} \end{align} $$ where $\varphi(u)=-\frac{\log(1-u)}{u(1-u)}\ge1$ on $(0,1)$.

Using $(1)$, $(2)$, $(3)$, and $(4)$, we can conclude that there is a $t$ that satisfies $(1)$ in the range $$ \frac{\log(n)}{n}\left(1-\frac{\log(\log(n))}{\log(n)}\right) \le t \le\frac{\log(n)}{n}\tag{5} $$ Comparing $(1)$ and $(3)$ gives $$ \begin{align} \frac{\log(\log(n)/n)}{\log(n)/n}-\frac{\log(t)}{t} &=n\frac{\log(\log(n))}{\log(n)}-\frac12+\frac1t\log\left(\frac{\sinh(t/2)}{t/2}\right)\\ &=\frac{1-\log(\xi)}{\xi^2}(\log(n)/n-t)\tag{6} \end{align} $$ for some $\xi$ between $\frac{\log(n)}{n}\left(1-\frac{\log(\log(n))}{\log(n)}\right)$ and $\frac{\log(n)}{n}$.

$(6)$ says that $$ \begin{align} \log(n)/n-t &=\frac{\xi^2}{1-\log(\xi)}\left(n\frac{\log(\log(n))}{\log(n)}-\frac12+\frac1t\log\left(\frac{\sinh(t/2)}{t/2}\right)\right)\\ &=\frac{\log(\log(n))}{n}+O\left(\frac{\log(\log(n))^2}{n\log(n)}\right)\tag{7} \end{align} $$ Since $x=e^{-t}$, we have that $x=1-t+O\left(\frac{\log(n)^2}{n^2}\right)$, and therefore, $$ \frac{n}{\log(\log(n))}\left(1-x-\frac{\log(n)}{n}\right) =-1+O\left(\frac{\log(\log(n))}{\log(n)}\right)\tag{8} $$ The desired conclusion follows from $(8)$.

Note that we avoided using big-O notation until $(7)$, where it was almost necessitated by the use of the mean value theorem.

Closer Investigation:

Comparison of the errors in $(3)$ and $(4)$ yield that $\frac{\log(n)}{n} \left(1-\frac{\log(\log(n))}{\log(n)}\right)$ is $\log(n)$ times closer to $t$ than is $\frac{\log(n)}{n}$. Thus, interpolation yields that a more accurate estimate would be $$ t\doteq\frac{\log(n)}{n}\left(1-\frac{\log(\log(n))}{1+\log(n)}\right)\tag{9} $$ Using $(9)$, we get that $$ \frac{n}{\log(\log(n))}\left(1-x-\frac{\log(n)}{n}\right)+1 \doteq\frac1{1+\log(n)}\tag{10} $$