Return to Answer

$\quad$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 $a_n=b_n^{-\beta}$.$\quad\square$

$\quad$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 $a_n=b_n^{-\beta}$.$\quad\square$

$\quad$Now $f(x)=x-\frac{r-2}{2r-2}x^2+O(x^3)$, derived from $$1-\theta_{n+1}=\left(1-\frac{\theta_n}{r-1}\right)^{r-1},$$ implies that $$a_n\sim \frac{2r-2}{r-2}\cdot\frac{1}{n},n\to \infty.$$

$\quad$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 $D=\frac{\alpha(\alpha+1)}{2}A^2-B\alpha$.

$\quad$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 $a_n=b_n^{-\beta}$.$\quad\square$

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 $C=A\alpha,\beta=\frac{1}{\alpha}$.

$\quad$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 $D=\frac{\alpha(\alpha+1)}{2}A^2-B\alpha$.