Calculate the limit of two interrelated sequences? 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, $\displaystyle\frac{1+2a+ab}{b-a}$.

It's obvious that $a_n$ and $b_n$ are in the same situation, so their limits highly depend on the initial values. Following are some points we can obtain from $a_1=1$ and $b_1=2$:


*

*$a_n>0$ and $b_n>0\ ;a_{n+1}-a_{n}=\frac{1+a_n}{b_n}>0$ and similarly $b_{n+1}-b_n>0$. Therefore, $\{a_n\}$ and $\{b_n\}$ are strictly increasing sequences;

*$b_{n+1}-a_{n+1}=\frac{(b_n-a_n)+(b_n^2-a_n^2)+a_n\cdot b_n(b_n-a_n)}{a_n\cdot b_n}$, and thus by induction $b_n>a_n$ for every $n$;

*$b_{n+1}-b_{n}=\frac{1+b_n}{a_n}>\frac{b_n}{a_n}>1$, which implies $b_n$ increases to $+\infty$;

*$\frac{a_{n+1}}{a_n}=1+\frac{1}{b_n}+\frac{1}{a_n\cdot b_n}$ converges to $1$ as $n\to \infty$.


Now we prove $\lim a_n$ exists and find its closed form. To show the existence, it suffices to show $\{a_n\}$ is bounded. First assume $a_n$ increases to infinity, and we will derive a contradiction with the last point listed above.
From the fact 
$$b_n(a_{n+1}+1)=(1+a_n)(1+b_n)=a_n(b_{n+1}+1)$$
Denote $c_n:=a_n+1$ and $d_n:=b_n+1$, then we obtain
$$
\begin{cases}
\frac{c_n-1}{c_n}=\frac{d_n}{d_{n+1}}\\
\frac{d_n-1}{d_n}=\frac{c_n}{c_{n+1}}
\end{cases}
\Rightarrow 
\begin{cases}
\frac{d_{n+1}}{d_n}=\frac{c_n}{c_n-1}\\
\frac{c_{n+1}}{c_n}=\frac{d_n}{d_n-1}
\end{cases}
$$
For $n\ge 2$,
$d_n=d_1\cdot \frac{d_2}{d_1}\cdots \frac{d_n}{d_{n-1}}=d_1\cdot \frac{c_1}{c_1-1}\cdots \frac{c_{n-1}}{c_{n-1}-1}$, which implies
$$
d_1(1-\frac{c_n}{c_{n+1}})=(1-\frac{1}{c_1})\cdots (1-\frac{1}{c_{n-1}})
$$
In fact,$$\frac{c_{n+1}}{c_n}=\frac{d_n}{d_n-1}=\frac{d_1\cdot \frac{c_1}{c_1-1}\cdots \frac{c_{n-1}}{c_{n-1}-1}}{d_1\cdot \frac{c_1}{c_1-1}\cdots \frac{c_{n-1}}{c_{n-1}-1}-1} \Rightarrow \frac{c_{n}}{c_{n+1}}=1-\frac{1}{d_1}((1-\frac{1}{c_1})\cdots (1-\frac{1}{c_{n-1}}))$$
Together with $d_1(1-\frac{c_{n+1}}{c_{n+2}})=(1-\frac{1}{c_1})\cdots (1-\frac{1}{c_{n}})$, we get $(1-\frac{1}{c_n})(1-\frac{c_n}{c_{n+1}})=1-\frac{c_{n+1}}{c_{n+2}}$. That is
$$
\frac{c_{n+1}}{c_{n+2}}-\frac{c_{n}}{c_{n+1}}=\frac{1}{c_{n}}-\frac{1}{c_{n+1}}
$$
Hence, for $n\ge 2$
$$
\frac{c_{n}}{c_{n+1}}-\frac{c_2}{c_3}=\frac{1}{c_2}-\frac{1}{c_{n}} $$
which is equivalent to
$$\frac{c_{n}}{c_{n+1}}+\frac{1}{c_{n}}=c(constant):=\frac{c_2}{c_3}+\frac{1}{c_2}=\frac{7}{6} \\ (c_1=2,d_1=3;c_2=3,d_2=6;c_3=\frac{18}{5})
$$
Now it is clear that $$\frac{c_{n+1}}{c_{n}}=\frac{c_n}{\frac{7}{6}c_n-1}=\frac{1}{\frac{7}{6}-\frac{1}{c_n}} $$
Well, the problem has been reduced to solve $c_n(=a_n+1)$, and it's you can use the same method to solve $b_n$, and I would like to leave this open to you, but note as I mentioned before if $a_n\to +\infty$, then $c_n\to +\infty$ and $$
\frac{c_{n+1}}{c_{n}}(=\frac{\frac{a_{n+1}}{a_n}+\frac{1}{a_n}}{1+\frac{1}{a_n}})=\frac{1}{\frac{7}{6}-\frac{1}{c_n}} \text{converges to } \frac{6}{7} \text{instead of } 1
$$ this contradicts the last point.
