TITLE: How prove $\lim_{n\to \infty}\frac{n}{\ln{(\ln{n}})}\left(1-a_{n}-\frac{n}{\ln{n}}\right)=-1$?
QUESTION [8 upvotes]: 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

REPLY [5 votes]: 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
$$
\frac{n}{\log(\log(n))}\left(1-x-\frac{\log(n)}{n}\right)+1
\doteq\frac1{1+\log(n)}\tag{10}
$$