Let the sequence $u_1, u_2, \ldots$ satisfy $u_{n+1} = u_n - u_n^2 + O(u_n^3)$.  Then it can be shown that if $u_n \to 0$ as $n \to \infty$, then $u_n = n^{-1} + O(n^{-2} \log n)$.  (See N. G. de Bruijn, <i>Asymptotic methods in analysis</i>, Section 8.5.)

This can be used to obtain asymptotics for $v_{n+1} = Av_n - Bv_n^2 + O(v_n^3)$, where $A$ and $B$ are constants.  Let $w_n = A^{-n} v_n$; this gives
$$ A^{n+1} w_{n+1} = A^{n+1} w_n - B A^n w_n^2 + O(A^n w_n^3)$$
and so
$$ w_{n+1} = w_n - BA^{-1} w_n^2 + O(w_n^3). $$
Then let $w_n = Ax_n/B$ to get
$$ Ax_{n+1}/B = Ax_n/B - B/A \cdot (Ax_n/B)^2 + O(x_n^3) $$
and after simplifying
$ x_{n+1} = x_n - x_n^2 + O(x_n^3)$.  This satisfies the initial requirements for $u_n$ (with some checking of the side condition); then substitute back.

But say I actually know that $u_{n+1} = P(u_n)$ for some polynomial $P$, with $P(z) = z - z^2 + a_3 z^3 + \cdots + a_d z^d.$   In this case it seems like it should be possible to get more explicit information about $u_n$.  Is there a known algorithm for computing an asymptotic series for $u_n$ as $n \to \infty$?