<s>[INST] Determine whether a recurrence relation converges and, if yes, find its limit. I have the sequence defined as $$ a_{n+1} = \dfrac{5a_n -6}{a_n -2} $$ for $n \geq 2$, $a_n \neq 2$ for all $n \geq 1$ and $a_n$ is a real number.
I want to determine whether the sequence converges and if it does, find it's limit.
There are no initial conditions given and also I can't think of any way to solve the recurrence relation.
One thing I considered is the following:
If the sequence converges then
$$ \lim a_{n+1} =\lim a_n = L $$
Then substitute in the recurrence relation to get a polynomial whose solutions are $a=1$,or $a=6$ and take cases for $a_1$, but I couldn't end up with the solution.
Finally I couldn't find a pattern for the sequence either so I am stuck. How can I solve this?
 [/INST] Let's experiment with some values of $n$.
\begin{align}
a_2 &= \frac{5a_1 - 6}{a_1 - 2} \\
a_3 &= \frac{19a_1 - 18}{3a_1 - 2} \\
a_4 &= \frac{77a_1 - 78}{13a_1 - 14} \\
a_5 &= \frac{307a_1 - 306}{51a_1 - 50}.
\end{align}
The coefficient of $a_1$ in the numerator generates the sequence $5, 19, 77, 307, \dots$ and the coefficient of $a_1$ in the denominator generates the sequence $1,3,13,51,\dots$.
Notice that
\begin{align}
5\cdot 4 - 1 &= 19 \\
19 \cdot 4 + 1 &= 77 \\
77 \cdot 4 - 1 &= 307 \\
\vdots
\end{align} and
\begin{align}
1\cdot 4 - 1 &= 3 \\
3 \cdot 4 + 1 &= 13 \\
13 \cdot 4 - 1 &= 51 \\
\vdots
\end{align} Let's express one of the terms from one of our sequences, say $307$, using the above pattern $$307 = ((5 \cdot 4 - 1)(4) + 1)(4) - 1 = 5 \cdot 4^3 - 4^2 + 4^1 - 4^0.$$ Given that $307$ is the coefficient of $a_1$ in the numerator of the expression for $a_5$, it looks like the coefficient of $a_1$ in the numerator of the expression for $a_n$ is given by
\begin{align}
\alpha_n :&= 5\cdot 4^{n-2} + \sum_{k = 0}^{n - 3}(-1)^{n-k}4^k = \frac{6\cdot 4^{n-1} + (-1)^n}{5}.
\end{align} Similarly, it looks like the coefficient of $a_1$ in the denominator of the expression for $a_n$ is given by $$\beta_n := \frac{4^{n-1} + (-1)^n}{5}.$$ The constant in the numerator and the denominator appear to alternate between the negative of one greater than and one less than the coefficient of $a_1$ in the numerator and the denominator, respectively, and so we conjecture that
\begin{align}
a_n &= \frac{\alpha_n a_1 - (\alpha_n + (-1)^n)}{\beta_n a_1 - (\beta_n + (-1)^n)} = \frac{(6\cdot 4^{n-1} + (-1)^n)(a_1 - 1) - 5(-1)^n}{(4^{n-1} + (-1)^n)(a_1 - 1) - 5(-1)^n}
\end{align} and prove by induction.
Keep in mind that $\beta_n a_1 - (\beta_n + (-1)^n)$ cannot equal zero and, hence, $a_1$ cannot equal $\frac{\beta_n + (-1)^n}{\beta_n}$ (e.g., $n = 2$ $\implies$ $a_1$ cannot equal $2$ and $n = 3$ $\implies$ $a_1$ cannot equal $\frac{2}{3}$).
In summary, if $a_1 = 1$, then $a_n = 1$ for all $n \geq 1$ and if $a_1 \neq 1$ and $a_1 \neq \frac{\beta_n + (-1)^n}{\beta_n}$, then $a_n \rightarrow 6$ as $n \rightarrow \infty$.
</s>