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, $\displaystyle\frac{1+2a+ab}{b-a}$.
 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.
A: 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 $lim_{n\rightarrow\infty} a_{n} = 5$ 
A: The answer is $a_n \to 5$ , $b_n \to \infty$.
I'm trying to prove that and I will edit this post if I figure out something.
EDIT:
I would write all this in comment instead in answer, but I cannot find how to do it.. maybe I need to have more reputation to do this (low reputation = low privileges:P)
Anyway, I still didn't solved it, but maybe something of that will help you. I will edit it when I think something out. 
EDIT:
After many transformations and playing with numbers, I think that the limit, for $a<b$, is $$ \frac{ab + 2a +1}{b-a}$$
But still cannot prove it.
(In statement above: $a = a_1 $ , $ b = b_1 $)
A: To proceed
Actually we can compute $\lim a_n$ explicitly, since
$$
\frac{c_{n+1}}{c_n}=\frac{1}{c-\frac{1}{c_n}} \text{ converges to } 1 \quad \Leftrightarrow \quad c_n \text{ converges to } \frac{1}{c-1}  $$
which is also equivalent to that
$ a_n \text{ converges to } \frac{1}{c-1}-1 $,
where $c=\frac{c_2}{c_3}+\frac{1}{c_2}$. 
We know$\ c_2=a_2+1=\frac{(1+a_1)(1+b_1)}{b_1},c_3=a_3+1=\frac{(+a_2)(1+b_2)}{b_2}=\frac{(1+a_1)^2(1+b_1)^2}{b_1(1+b_1+a_1b_1)}$ 
Now we get $c=\frac{1+2b_1+a_1b_1}{(1+a_1)(1+b_1)}$ and 
$$
a_n\to \frac{1}{c-1}-1=\frac{1+2a_1+a_1b_1}{b_1-a_1}
$$
Edit:
Thanks to Mihai Dicu, once we notice that $\frac{1}{1+a_{n+1}}-\frac{1}{1+b_{n+1}}=\frac{1}{1+a_{n}}-\frac{1}{1+b_{n}}=\frac{1}{1+a_{1}}-\frac{1}{1+b_{1}}$, it is rather easy to find the limit. From my previous answer, if $b_1>a_1>0$, we can actually show that both $a_n$ and $b_n$ are stictly increasing, and $b_n\to +\infty$. Therefore,
$$
\frac{1}{1+a_n}=\frac{1}{1+b_n}+\frac{1}{1+a_1}-\frac{1}{1+b_1}>\frac{1}{1+a_1}-\frac{1}{1+b_1}>0 $$
which shows that ${a_n}$ is bounded, and thus its limits exists.
$$
\lim_{n\to \infty}b_n=+\infty \Rightarrow \lim_{n\to \infty}\frac{1}{1+a_n}=\frac{1}{1+a_1}-\frac{1}{1+b_1}=\frac{b_1-a_1}{(1+a_1)(1+b_1)}
$$
which is equivalent to 
$$
\lim_{n\to \infty}a_n=\frac{(1+a_1)(1+b_1)}{b_1-a_1}-1=\frac{1+2a_1+a_1b_1}{b_1-a_1}
$$
