$\quad$Restate what was said in my comments. Using a computer to view my answer will be better.

**Theorem.** Let $A>0,\alpha>0$ and $f(x)$ be the function satisfying the following condition
$$f(x)=x-Ax^{1+\alpha}+Bx^{1+2\alpha}+O\left(x^{1+3\alpha}\right),\ x\to 0.$$
If $a_{n+1}=f(a_n)$ and the sequence $(a_n)_{n\geqslant 1}$ is monotone decreasing to zero, then
$$a_n=\frac{1}{C^{\beta}n^{\beta}}+\left(B-\frac{(\alpha+1)A^2}{2}\right)\frac{\log n}{C^{2+\beta}n^{1+\beta}}+O\left(n^{-1-\beta}\right)$$
holds as $n\to \infty$, where $C=A\alpha,\beta=\frac{1}{\alpha}$.

**Proof.** Putting $b_n=a_n^{-\alpha}$, we have $b_n\to +\infty$ and
\begin{align*}
  b_{k+1}& =\frac{1}{a_{k+1}^{\alpha}}=b_k\left(\frac{a_{k+1}}{a_k}\right)^{-\alpha}\\
  & =b_k\left(1-Aa_k^{\alpha}+Ba_k^{2\alpha}+O\left(a_k^{3\alpha}\right)\right)^{-\alpha}\\
  & =b_k\left(1-\frac{A}{b_k}+\frac{B}{b_k^2}+O\left(b_k^{-3}\right)\right)^{-\alpha}\\
  & =b_k+C+\frac{D}{b_k}+O\left(b_k^{-2}\right),\tag{1}
\end{align*}
where $D=\frac{\alpha(\alpha+1)}{2}A^2-B\alpha$.

The O'Stolz Theorem shows that
$$\lim_{n\to \infty}\frac{b_n}{n}=\lim_{n\to \infty}(b_{n+1}-b_n)=C.$$
Sum the formula (1) from $k=1$ to $k=n-1$ and get
\begin{align*}
  b_n& =b_1+C(n-1)+\sum_{k=1}^{n-1}\frac{D+o(1)}{Ck}+O\left(\sum_{k=1}^n\frac{1}{k^2}\right)\\
  & =Cn+\frac{D+o(1)}{C}\log n+O(1)\\
  & =Cn\left(1+O\left(\frac{\log n}{n}\right)\right).
\end{align*}
Sum again and we get
\begin{align*}
  b_n& =b_1+C(n-1)+\sum_{k=1}^{n-1}\frac{D+O\left(\frac{\log k}{k}\right)}{Ck}+O\left(1\right)\\
  & =Cn+\frac{D}{C}\log n+O(1).
\end{align*}
The proof completes with the substitution $a_n=b_n^{-\beta}$.$\quad\square$