Finding the equivalent of a polynomial sequence 
*

*Let $P_n=X^n+X-1$ be a polynomial with only one positive real root
$x_n$.
The sequence $(x_n)$ is obviously well defined and it's
not hard to find that $\lim\limits_{n \to +\infty} x_n=1$
In fact $(x_n)$ is an increasing bounded sequence. Let $l$ be its limit. We obviously have $l \le 1$.
As $P_n$ is increasing on $\mathbb{R_+}$,
$0=x_n^n+x_n-1\le l^n+l-1$ so the limit when $n \to \infty$ satisfies $l-1\ge 0$.
So $l=1$.
Then I need to find an equivalent sequence to $x_n-1$.
What I did:
We have $n=\frac{\ln(1-x_n)}{\ln x_n}$.
I therefore tried to find an equivalent of the form $n^{\alpha}(\ln n)^{\beta}$ and after some trials I found that $-\frac{\ln n}{n}$ works.
Since
$\displaystyle\frac{n}{\ln n} (1-x_n)=\displaystyle\frac{\frac{\ln(1-x_n)}{\ln x_n} (1-x_n)}{\ln\frac{\ln(1-x_n)}{\ln x_n}}$ we only need to show that $\displaystyle\lim\limits_{n \to +\infty}\frac{\ln(1-x_n)}{\ln\frac{\ln(1-x_n)}{\ln x_n}}=-1$
because $\lim\limits_{n \to +\infty}\frac{(1-x_n)}{\ln x_n}=-1$ (limit of the tangency point of $\ln$ at $1$).
we can show that $\displaystyle\lim\limits_{x \to 1^{-}}\frac{\ln(1-x)}{\ln\frac{\ln(1-x)}{\ln x}}=-1$
To simplify it, I noticed that both numerator and denominator tend to $\pm \infty$ when $x\to 1^-$.
So we can apply L'Hôpital's Rule and then :
$\frac{d}{dx} \ln(1-x)=\frac1{x-1}$
and $\frac{d}{dx}\left( \ln\frac{\ln(1-x)}{\ln x} \right)=\frac{x \ln x +(1-x)\ln{(1-x)}}{x(x-1) \ln x \ln(1-x)}$
the limit becomes $\frac{x \ln x +(1-x)\ln{(1-x)}}{x\ln x \ln(1-x)}=\frac 1 {\ln(1-x)}+\frac{(1-x)}{x\ln x} \to_{x \to 1^-} 0+(-1)=-1$.
Finally we have that $\frac{n}{\ln n} (1-x_n) \to -1$ and then $x_n-1 \sim -\frac{\ln n}{n}$
My questions:

*

*Is this correct ?

*Is there a better way to proceed ? Maybe with the MVT ?

Thanks in advance.
 A: I'll ignore the subscript for readability. Write $x = 1 - \frac{c}{n}$. We have $x^n = 1 - x$ which gives
$$\left( 1 - \frac{c}{n} \right)^n = \frac{c}{n}.$$
Taking logarithms gives
$$n \log \left( 1 - \frac{c}{n} \right) = \log c - \log n.$$
Approximating the LHS as $-c + O \left( \frac{c^2}{n} \right)$ gives $c + \log c = \log n + O \left( \frac{c^2}{n} \right)$ which gives $c \approx \log n - \log \log n$; estimating the error in this approximation involves estimating
$$\begin{eqnarray*} \log (\log n - \log \log n) &=& \log \left( \log n \left( 1 - \frac{\log \log n}{\log n} \right) \right) \\ &=& \log \log n + \log \left( 1 - \frac{\log \log n}{\log n} \right) \\ &=& \log \log n - \frac{\log \log n}{\log n} + O \left( \left( \frac{\log \log n}{\log n} \right)^2 \right) \end{eqnarray*}$$
which gives something like
$$c = \log n - \log \log n + \frac{\log \log n}{\log n} + O \left( \left( \frac{\log \log n}{\log n} \right)^2 \right).$$
Note that this error term is much larger than $O \left( \frac{c^2}{n} \right) = O \left( \frac{(\log n)^2}{n} \right)$; it decays very slowly. You can get more terms out of known asymptotics for the Lambert W function since we have $ce^c \approx n$ so $c \approx W(n)$.
To give a flavor of the accuracy here, when $n = 100$ we have (according to WolframAlpha)
$$x_{100} \approx 0.966584$$
and hence $c_{100} = 100(1 - x_{100}) \approx 3.342$. The approximation above for $c_{100}$ gives, taking one term at a time,
$$\log 100 \approx 4.605$$
$$\log 100 - \log \log 100 \approx 3.078$$
$$\log 100 - \log \log 100 + \frac{\log \log 100}{\log 100} \approx 3.410.$$
So taking the first term is not that great, and even taking the first three terms only gives us a single digit of accuracy. We also have $W(100) \approx 3.386$ which is at least two digits of accuracy. We could get more accuracy by starting with $W(n)$ and expanding from there, probably.
