How to evaluate $\lim_{n\to\infty} a_n$ If $a_1=1$ and $a_{n+1}=\frac{4+3a_n}{3+2a_n}$,$n\geq1$, then how to prove that $a_{n+2}>a_{n+1}$ and if $a_n$ has a limit as ${n\to\infty}$ then how to evaluate $\lim_{n\to\infty} a_n$ ?
 A: Using deepak's hint:
$$\lim_{n\to \infty}\frac{4+3a_n}{3+2a_n}=\frac{4+3\lim_{n\to \infty}a_n}{3+2\lim_{n\to \infty}a_n}=L$$
This becomes 
$\frac{4+3L}{2L+3}=\lim_{n\to \infty} a_n=L$$\implies 2L^2-4=0$$$\implies L=\pm\sqrt{2}$$
Since the sequence is clearly increasing,the positive value of L will be considered.
Please point out if I have made some mistake
A: Consider the function $f(x) = \frac{4+3x}{3+2x} = \frac32 - \frac{1}{2(3+2x)}$. It is easy to check.


*

*$f(x)$ is a strictly increasing function from $[0,\infty)$ to $[0,\infty)$.

*$\sqrt{2}$ is a fixed point for $f(x)$.


So start from any number $a \in (0,\sqrt{2})$, if we construct a sequence $a_n$ by
$$a_{n} = \begin{cases} a, & n = 1\\ f(a_{n-1}), & n > 1\end{cases}$$
We will have
$$a_1 < \sqrt{2} 
\implies a_2 = f(a_1) < f(\sqrt{2}) = \sqrt{2}
\implies a_3 = f(a_2) < f(\sqrt{2}) = \sqrt{2}
\implies \cdots$$
This means all $a_n < \sqrt{2}$. By setting $a = 1$, we will recover the 
sequence $a_n$ we have. Notice
$$a_1 = 1 \quad\text{ and }\quad a_2 = f(a_1) = f(1) = \frac75 > a_1$$
The strictly increasing property of $f(x)$ leads to
$$a_1 < a_2 \implies a_2 = f(a_1) < f(a_2) = a_3 \implies a_3 = f(a_2) < f(a_3) = a_4 \implies \cdots$$
This means the sequence we have is monotonic increasing.
Combine these two observation, $a_n$ is a monotonic increasing sequence, bounded from above by $\sqrt{2}$. So it converges to some limit $a_\infty \le \sqrt{2}$. Since $f(x)$ is continuous, we have
$$f(a_\infty) = f(\lim_{n\to\infty} a_n) = \lim_{n\to\infty} f(a_n) = \lim_{n\to\infty} a_{n+1} = a_\infty$$
So $a_\infty$ is a fixed point of $f(x)$. Solving the equation $f(x) = x$ for $x \in [1,\sqrt{2}]$ leads to $a_\infty = \sqrt{2}$.
For this particular problem, there is actually a way to compute $a_n$ explicitly.
Rewrite $a_n$ as $\frac{p_n}{q_n}$ for some $p_n, q_n$ to be determined. We can rewrite the recurrence relation as
$$a_{n+1} = \frac{3a_n + 4}{2 a_n + 3} \quad\iff\quad
\frac{p_{n+1}}{q_{n+1}} = \frac{3p_n + 4 q_n}{2 p_n + 3q_n}\tag{*1}$$
If we scale $p_n, q_n$ so that they satisfies following matrix equation
$$\begin{bmatrix}p_{n+1}\\ q_{n+1}\end{bmatrix}
= \begin{bmatrix}3 & 4\\ 2 & 3\end{bmatrix}
 \begin{bmatrix}p_{n}\\ q_{n}\end{bmatrix}$$
It will solve the RHS of $(*1)$. What this means is if we define two sequences
$p_n, q_n$ by
$$
\begin{bmatrix}p_{n}\\ q_{n}\end{bmatrix} =
\begin{bmatrix}3 & 4\\ 2 & 3\end{bmatrix}^{n-1}
 \begin{bmatrix}1\\1\end{bmatrix}\quad\text{ for } n \ge 1$$
Our $a_n$ will be equal the quotient of these two pair of sequences. Diagonalizing the matrix and with a little bit of algebra, one find
$$a_n = \sqrt{2}\left[\frac{
(\sqrt{2}+1)^{2n-1} - (\sqrt{2}-1)^{2n-1}
}{
(\sqrt{2}+1)^{2n-1} + (\sqrt{2}-1)^{2n-1}
}\right]$$
Using this representation, it is immediately clear $\lim_{n\to\infty} a_n = \sqrt{2}$.
