Roots of $X^n-nX+1$ asymptotic expansion For all $n \geq 2$, let's consider $P_n = X^n-nX+1$. Using Rolle theorem we can then show that $P_n$ has a single root $u_n$ in $]0,1[$.
I would like to get an asymptotic expansion with $4$ terms of $u_n$.
First I need to find the limit of $u_n$ so we have :
$u_n^n -nu_n+1 = 0$ thus : $l^n/n - l = -1/n$ so $l = 0$ if we assume that $u_n$ has a limit.
Yet from now I don't really know how to proceed.
 A: If this can help:
A first approximation is $x_n=\dfrac1n$. A better approximation can be found in the form $\dfrac{1+t}n$. We write
$$\left(\frac{1+t}n\right)^n-n\left(\frac{1+t}n\right)+1=0,$$
$$\left(\frac{1+t}n\right)^n\approx\frac{1+nt}{n^n}=t,$$
$$t\approx \frac1{n^n-n},$$
and
$$x_n\approx\frac{1+\dfrac1{n^n-n}}n.$$
E.g., $x_5\approx0.2000641025641$ and $x_5^5-5x_5+1\approx3.28\cdot10^{-10}$
That process can be continued.

Another method is to write the Newton's iterates, starting from the initial approximation. Accuracy is very high, but the expressions quickly grow.
A: Hopefully I didnt do any mistake:
Note that
$$P_n(\frac{1}{n}+\frac{1}{n^{n+1}})= (\frac{1}{n}+\frac{1}{n^{n+1}})^n - \frac{1}{n^n}>0$$
Now, let $\alpha >1$. We have
$$P_n(\frac{1}{n}+\frac{\alpha}{n^{n+1}}) <0 \Leftrightarrow \\
\frac{1}{n}+\frac{\alpha}{n^{n+1}} < \frac{\sqrt[n]{\alpha}}{n} \Leftrightarrow \\
n(\sqrt[n]{\alpha} -1) \geq \frac{\alpha}{n^{n-1}}$$
Now,
$$\lim_n n(\sqrt[n]{\alpha} -1) =\lim_n \frac{\alpha^\frac{1}{n}-\alpha^0}{\frac{1}{n}-1}=\ln(\alpha) >0$$
and
$$\lim_n \frac{\alpha}{n^{n-1}}=0$$
This shows that for all $\alpha >1$ there exists some $N$ so that, for all $n>N$ we have
$$P_n(\frac{1}{n}+\frac{\alpha}{n^{n+1}}) <0 $$
It follows that asymptotically, we have
$$\frac{1}{n}+\frac{1}{n^{n+1}} < u_n < \frac{1}{n}+\frac{\alpha}{n^{n+1}} \qquad \forall \alpha >1$$
[i.e. For each $\alpha >1$, there exists a $N$ such that the above holds for all $n>N$]
A: Here's an idea to get a complete asymptotics (obviously not in fixed powers of $n$, too).
For a fixed $n>1$, the solution $w=w_n(z)$ of $w=1+zw^n$ has a known power series $$w_n(z)=\sum_{k=0}^\infty\binom{nk}{k}\frac{z^k}{(n-1)k+1}$$ (a way to get it is basically Lagrange's inversion theorem). Thus, if $v_n=nu_n$ for our $u_n$, then $$v_n=1+n^{-n}(v_n)^n\implies v_n=w_n(n^{-n})\implies u_n=n^{-1}w_n(n^{-n}).$$
