We have the sequence $$a_1=\frac{9}{4} \quad \qquad a_{n+1}=2\frac{a_n^3-a_n^2+1}{3a_n^2-4a_n+1}$$ that converges to $\frac{9}{4}$.
I want to check how fast the convergence is.
Do we have to calculate $\lim_{n\rightarrow \infty}\frac{|a_{n+1}-9/4|}{|a_n-9/4|}$ ?
Best Answer
Hint: Define $$f(x)=2\frac{x^3 - x^2 + 1}{3 x^2 - 4 x + 1}$$
Prove the sequence $a_{n+1}=f(a_n)$ converge to $2$ with $a_1 =\frac{9}{4}$ as you said.
\begin{align} a_{n+1} &= f(a_n) \\ &= 2\frac{(a_n-2)((a_n-2)^2+1)}{((a_n-2)+1)(3(a_n-2)+5)} \tag{1}\\ \end{align}
Denote $b_n = a_n-2$, we have $b_n \rightarrow 0$ when $n\rightarrow +\infty$ or $b_n=\mathcal{o}(1)$. And from (1) we have
$$b_{n+1}+2 = f(b_n+2)$$ or \begin{align} b_{n+1} &=f(b_n+2)-2 \\ &= \frac{b_n(b_n^2+1)}{(b_n+1)(3b_n+5)} -2 \\ &= \frac{2}{5}b_n(b_n^2+1)(1-b_n+\mathcal{O}(b_n^2))(1-\frac{3}{5}b_n+\mathcal{O}(b_n^2)) -2\\ &= \frac{4}{5}b_n^2-\frac{22}{5}b_n^3+\mathcal{O}(b_n^4)\\ &= \frac{4}{5}b_n^2(1+\mathcal{O}(b_n)) \tag{2}\\ \end{align} Then, $$\ln(b_{n+1})= \ln(\frac{4}{5})+ 2\ln(b_{n}) +\ln(1+\mathcal{O}(b_n)) $$ $$\iff (\ln(b_{n+1})-\ln(\frac{4}{5}))= 2(\ln(b_{n})-\ln(\frac{4}{5})) +\mathcal{O}(b_n) \tag{3}$$ We notice that from (2), we can deduce also that $b_{n+1}=\mathcal{O}(b_n^2)=\mathcal{o}(b_n)$ or $\mathcal{O}(b_{n+1})=\mathcal{o}(b_n)$. So, $$(3)\iff (\ln(b_{n+1})-\ln(\frac{4}{5})+\mathcal{O}(b_{n+1}))= 2(\ln(b_{n})-\ln(\frac{4}{5})+\mathcal{O}(b_n)) $$ $$\iff \ln(b_{n})-\ln(\frac{4}{5})+\mathcal{O}(b_{n})= 2^{n-1}(\ln(b_{1})-\ln(\frac{4}{5})) =2^{n-1}\ln(\frac{5}{16}) $$ or $$b_n=\frac{4}{5} \left(\frac{5}{16} \right)^{2^{n-1}}$$ (because $b_n \rightarrow 0$ when $n \rightarrow +\infty$ then $\exp(\mathcal{O}(b_{n}))=1$)
Conclusion: $$a_n \approx 2+\frac{4}{5} \left(\frac{5}{16} \right)^{2^{n-1}}$$