Let $0<a<1$ and $a_{n+1}=2-\frac{1-a}{a_n}$ a recursive sequence and $a_1 \neq0$.

Find the limit of the sequences if it exists.

The fixed points of the sequence are $1-\sqrt{a},1+\sqrt{a}$

I managed to prove that if $a_1 \in I=(1-\sqrt{a},1+\sqrt{a})$ then $a_n$ is increasing and $a_n \to 1+\sqrt{a}$

What can i say when $a_1 \notin I$?

I believe that if $a_1 \notin I$ then $a_n$ does not converge to a real number,but i do not see how to prove it.

Can someone help me with this?

Thank you in advance.

  • $\begingroup$ How do you fix $a_0$? $\endgroup$ – Paolo Leonetti Jan 13 '18 at 0:35
  • $\begingroup$ $a_1$ which the initial term ,is not fixed..i have to study the limiting behavior of the sequence by considering cases for $a_1$ $\endgroup$ – Marios Gretsas Jan 13 '18 at 0:38
  • $\begingroup$ So, do you want an answer depending on both $a \in (0,1)$ and $a_1 \in \mathbf{R}$? (I suppose $a_1 \neq 0$..) $\endgroup$ – Paolo Leonetti Jan 13 '18 at 0:39
  • $\begingroup$ I want some help with the case i mentioned. $\endgroup$ – Marios Gretsas Jan 13 '18 at 0:40

Hint: Based on your proof of convergence for $a_1\in(1-\sqrt a, 1+\sqrt a)$ break the problem into few steps:

  1. $a_1 \geq 1+\sqrt a$; show that for all $n$, $a_n\geq1+\sqrt a$ as well. Then show that $a_n$ is decreasing by seeing that $a_{n+1}-a_n \leq 0 \\$.

  2. If $a_k$ is negative for some $k$, then $a_{k+1}$ and all the next terms jumps to step 1.

  3. If all $a_n$ get sandwich between $0$ and $1-\sqrt a$, what is the contradiction?!


If we define $b_1=1$ and $\displaystyle b_n=\prod^{n-1}_{k=1}a_k$, we see that $a_n=\dfrac{b_{n+1}}{b_n}$, and the recursion becomes $$b_{n+2}=2\,b_{n+1}-(1-a)\,b_n.$$ The general solution of this linear recursion is $$b_n=A\,(1-\sqrt{a})^n+B\,(1+\sqrt{a})^n,\tag1$$ where $A,B$ can be determined from the initial values $b_1=1$ and $b_2=a_1$. It is clear that the limit of $a_n$ will be $1+\sqrt{a}$ whenever $B\neq0$, since $(1-\sqrt{a})^n\to0$. But $B=0$ is possible only if $a_n=a_1=1-\sqrt{a}$. So the sequence is converging to $1+\sqrt{a}$ for all other starting values for which the sequence is defined, i.e. all except a countable number of starting values making one term $a_k=0$, starting values which can easily be determined from (1).


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