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Let $a_n$ and $b_n$ be two sequences defined by the recurrence relation \begin{align} a_{n + 1} = \frac{{1 + a_n + a_n b_n }}{{b_n }},\qquad b_{n + 1} = \frac{{1 + b_n + a_n b_n }}{{a_n }} \end{align} with $a_1=1$ and $b_1=2$. Find $\mathop {\lim }\limits_{n \to \infty } a_n$.


Simple manipulations yield that \begin{align} a_n = \frac{{1 + b_n }}{{b_{n + 1} - b_n }},\qquad b_n = \frac{{1 + a_n }}{{a_{n + 1} - a_n }} \end{align} I tried to find the first few terms of $a_n$ and $b_n$ but it seem that they doesn't have a general form. Once can see that both sequences increasing fast and randomly. Also, I tried to insert $b_n$ in $a_n$, but I got a complicated result and it doesn't help or indicate to anything.

Any help is appreciated. Thanks

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marked as duplicate by Martin R, rtybase, Sangchul Lee sequences-and-series Apr 21 '18 at 11:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Almost immediately you can deduce (with a bit of induction) that $a_{n+1}-a_n=\frac{1+a_n}{b_n}>0$ $\endgroup$ – rtybase Apr 21 '18 at 11:05
  • $\begingroup$ and $$b_{n+1}-b_n=\frac{1+b_n}{a_n}>0$$ $\endgroup$ – Dr. Sonnhard Graubner Apr 21 '18 at 11:11
  • $\begingroup$ yes, it's ok thanks all. $\endgroup$ – mwomath Apr 21 '18 at 11:14