Suppose that $a_1{\in}\mathbb{R}$ and $a_{n+1}=\dfrac{2a_n-1}{2-a_n}$ for all $n{\ge}1$, ($a_n{\neq}2$). Evaluate$\lim\limits_{n\to+\infty}a_n$.


From the recursion formula, it's esay to obtain $$\frac{1}{a_{n+1}+1}-\frac{1}{2}=3\left(\dfrac{1}{a_{n}+1}-\dfrac{1}{2}\right).$$ This shows that $\left\{\dfrac{1}{a_{n}+1}-\dfrac{1}{2} \right\}$ is a geometric sequence.Hence,$$\dfrac{1}{a_{n}+1}-\frac{1}{2}=\left(\dfrac{1}{a_{1}+1}-\dfrac{1}{2}\right)\cdot 3^{n-1}=\gamma \cdot 3^{n-1},$$where $\gamma= \dfrac{1}{a_{1}+1}-\dfrac{1}{2}$. Then,$$a_n=\frac{2}{2\gamma \cdot 3^{n-1}+1}-1.$$

  • $\textbf{Case 1}$

If $a_1=1$, Then $\gamma=0$. In this case, $a_n \equiv 1$ for all $n$. Thus, $\lim\limits_{n \to \infty}a_n=1$.

  • $\textbf{Case 2}$

If $a_1 \neq 1$, then $\gamma \neq 0$. Let $n \to \infty$. It follows that $$\lim_{n \to \infty}a_n=\lim_{n \to \infty}\left(\frac{2}{2\gamma \cdot 3^{n-1}+1}-1\right)=0-1=-1.$$ Summing up the above,$$\lim_{n \to \infty}a_n=\begin{cases}1, &a_1=1;\\-1,&a_1 \neq 1,a_n \neq 2.\end{cases}$$

Are there other solutions without finding general term?

  • $\begingroup$ With the general theory of fixed point iteration, it suffices to note that for the continuously differentiable function $f(x) = \frac{2x-1}{2-x}$, $x = -1$ is a fixed point and $|f'(-1)| < 1$ $\endgroup$ – Omnomnomnom May 31 '18 at 2:52
  • $\begingroup$ Along those lines, one approach you might like is that of applying the mean-value theorem to approximate $|a_{n+1} - a_n|$ $\endgroup$ – Omnomnomnom May 31 '18 at 2:53
  • $\begingroup$ @Omnomnomnom $x=1$ is also the fixed point. Besides,$f'(x)$ is not bounded over all $R$... $\endgroup$ – mengdie1982 May 31 '18 at 3:35
  • $\begingroup$ Two fixed points we would have are $a_n$ = 1 and -1 $\endgroup$ – Haran May 31 '18 at 5:55
  • $\begingroup$ Either the answer diverges in a non-cyclic way or it goes to 1 or -1. A cycle formation is not possible as it will lead to a quadratic, and we already have two solutions, namely 1 and -1. $\endgroup$ – Haran May 31 '18 at 6:04

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