# Find the limit of $a_{n + 1} = \frac{{1 + a_n + a_n b_n }}{{b_n }}$ [duplicate]

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

## marked as duplicate by Martin R, rtybase, Sangchul Lee sequences-and-series StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 21 '18 at 11:09

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