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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 $\lim_{n\rightarrow \infty}\frac{|a_{n+1}-9/4|}{|a_n-9/4|}$ ?

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    $\begingroup$ How do you define the speed of convergence of a sequence? $\endgroup$
    – José Carlos Santos
    Feb 12, 2021 at 10:29
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    $\begingroup$ I don't know. It depends upon the definition that you are working with. $\endgroup$
    – José Carlos Santos
    Feb 12, 2021 at 10:33
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    $\begingroup$ @MaryStar in computer science one way to do so is to see if $a_n=O(f(n))$ where $f(n)=n^\epsilon$ or $e^n$... $\endgroup$
    – Michelle
    Feb 12, 2021 at 10:34
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    $\begingroup$ Your question about the limit : If the convergence is linear, yes, but if for example the convergence is quadratic, you have to place an exponent $2$ on the denominator... $\endgroup$
    – Jean Marie
    Feb 12, 2021 at 10:37
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    $\begingroup$ Replace $a_n$ also to get $b_{n+1}=\frac{2{b_n}^2(b_n+2)}{(3b_n+5)(b_n+1)}$ and get $b_{n+1}\sim\frac 45{b_n}^2$ since $b_n\to 0$. $\endgroup$
    – zwim
    Feb 12, 2021 at 11:08

2 Answers 2

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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: $$a_n \approx 2+\frac{4}{5} \left(\frac{5}{16} \right)^{2^{n-1}}$$

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  • $\begingroup$ There is a bit of notation abuse on $b_n$ when the $o({b_n}^2)$ term is suddently ignored. You should call it $c_n$ or $B_n$ when going next line with the logarithm relation. $\endgroup$
    – zwim
    Feb 12, 2021 at 16:14
  • $\begingroup$ @zwim: you are right. Now, honestly I have a doubt about my answer :) $\endgroup$
    – NN2
    Feb 12, 2021 at 16:48
  • $\begingroup$ The answer still stands, but I agree it would be painful to justify it fully. Basically $\ln(b_{n+1})=\ln(4/5)+2\ln(b_n)+O(b_n)$ (actually $-11b_n/2$ if we develop the ln) and this $O(b_n)\to 0$ is still very negligible compared to $\ln(b_n)\to-\infty$ so we can ignore it. $\endgroup$
    – zwim
    Feb 12, 2021 at 17:26
  • $\begingroup$ @zwim: I modified the answer. Thank you! $\endgroup$
    – NN2
    Feb 12, 2021 at 18:04
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You want to find how many terms it takes in order for $|a_n-\frac{9}{4}|<\epsilon$, where $\epsilon$ is some chosen positive constant. For example, how many terms would it take for $a_n-\frac{9}{4}<10^{-3}$?

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