TITLE: Calculate the limit of two interrelated sequences?
QUESTION [8 upvotes]: I'm given two sequences:
$$a_{n+1}=\frac{1+a_n+a_nb_n}{b_n},b_{n+1}=\frac{1+b_n+a_nb_n}{a_n}$$
as well as an initial condition $a_1=1$, $b_1=2$, and am told to find: $\displaystyle \lim_{n\to\infty}{a_n}$.
Given that I'm not even sure how to approach this problem, I tried anyway. I substituted $b_{n-1}$ for $b_n$ to begin the search for a pattern. This eventually reduced to:
$$a_{n+1}=\frac{a_{n-1}(a_n+1)+a_n(1+b_{n-1}+a_{n-1}b_{n-1})}{1+b_{n-1}+a_{n-1}b_{n-1}}$$
Seeing no pattern, I did the same once more:
$$a_{n+1}=\frac{a_{n-2}a_{n-1}(a_n+1)+a_n\left(a_{n-2}+(a_{n-1}+1)(1+b_{n-2}+a_{n-2}b_{n-2})\right)}{a_{n-2}+(a_{n-1}+1)(1+b_{n-2}+a_{n-2}b_{n-2})}$$
While this equation is atrocious, it actually reveals somewhat of a pattern. I can sort of see one emerging - though I'm unsure how I would actually express that. My goal here is generally to find a closed form for the $a_n$ equation, then take the limit of it.
How should I approach this problem? I'm totally lost as is. Any pointers would be very much appreciated!
Edit:
While there is a way to prove that $\displaystyle\lim_{n\to\infty}{a_n}=5$ using $\displaystyle f(x)=\frac{1}{x-1}$, I'm still looking for a way to find the absolute form of the limit, $\displaystyle\frac{1+2a+ab}{b-a}$.

REPLY [2 votes]: From a previous answer (Coiacy) we know that $a_{n}$ is increasing and
 $lim_{n\rightarrow\infty} b_{n} = \infty$.
It is easy to prove equalities:
1) $ 1+a_{n+1}=\frac{(1+a_{n})(1+b_{n})}{b_{n}}$;
2) $1+b_{n+1}=\frac{(1+a_{n})(1+b_{n})}{a_{n}}; $
3) $\frac{1}{1+a_{n+1}}-\frac{1}{1+b_{n+1}}= \frac{1}{1+a_{n}}-\frac{1}{1+b_{n}}.$
From  3)  it follows that $\frac{1}{1+a_{n}}-\frac{1}{1+b_{n}} = \frac{1}{1+a_{1}}-\frac{1}{1+b_{1}} = \frac{1}{6}$ and $\frac{1}{1+a_{n}}= \frac{1}{6} + \frac{1}{1+b_{n}} > \frac{1}{6}$. 
From here we have that $ a_{n}< 5 $ which means that $a_{n}$ is monotone and bounded and $lim_{n\rightarrow\infty} a_{n} = l$ where $l\in (0, 5]$.
Because $lim_{n\rightarrow\infty} b_{n} = \infty$ it follows that $lim_{n\rightarrow\infty}\frac{1}{1+a_{n}}$ 
$= \frac{1}{6}+lim_{n\rightarrow\infty}\frac{1}{1+b_{n}} = \frac{1}{6}$  and consequently $lim_{n\rightarrow\infty} a_{n} = 5$