How to obtain general terms

Question

$$ a_{n+1} = \frac{3a_n -1 }{4a_n -1}, \qquad n=1,2,3 \ldots$$

Answer 1

$$a_2=\dfrac{3a_1-1}{4a_1-1}$$ and $$a_3=\dfrac{3\dfrac{3a_1-1}{4a_1-1}-1}{4\dfrac{3a_1-1}{4a_1-1}-1}=\frac{5a_1-2}{8a_1-3}$$

and more generally,

$$a_n=\frac{p_na_1-q_n}{r_na_1-s_n}.$$

Then

$$a_{n+1}=\dfrac{3\dfrac{p_na_1-q_n}{r_na_1-s_n}-1}{4\dfrac{p_na_1-q_n}{r_na_1-s_n}-1}=\frac{(3p_n-r_n)a_1-(3q_n-s_n)}{(4p_n-r_n)a_1-(4q_n-s_n)}$$

and we obtain a system of linear recurrences,

$$\begin{cases}p_{n+1}=3p_n-r_n,\\q_{n+1}=3q_n-s_n,\\r_{n+1}=4p_n-r_n,\\s_{n+1}=4q_n-s_n.\end{cases}$$

From the first and the third,

$$p_{n+2}=3p_{n+1}-r_{n+1}=3p_{n+1}-4p_n+r_n=3p_{n+1}-4p_n+3p_n-p_{n+1},$$ or

$$p_{n+2}-2p_{n+1}+p_n=0.$$

With the given initial conditions ($a_1=\frac{1\cdot a_1-0}{0\cdot a_1-(-1)}$), the solution is

$$p_n=2n-1,$$ which implies $$r_n=4n-4.$$

The resolution is similar for $q_n,s_n$.

$$a_n=\frac{(2n-1)a_1-(n-1)}{(4n-4)a_1-(2n-3)}.$$

In the limit,

$$a_\infty=\frac12$$ as can be directly verified.

Answer 2

Suppose $b_{n} = 4a_n - 1$. Then $b_{n + 1} = 2 - \frac{1}{b_n}$. That means, that if $c_n = (-1)^n b_n$, then $c_{n + 1} = 2(-1)^n + \frac{1}{b_n}$, thus $c_n$ can be expressed via continued fractions as $[\{(-1)^k2\}_{k = n}^1, c_0] = -2 + \frac{1}{2 + \frac{1}{-2 + \frac{1}{2 + \frac{1}{\ddots + \frac{1}{c_0}}}}}$. Thus

$$a_n = \frac{(-1)^n[\{(-1)^k2\}_{k = n}^1, (-1)^n(4a_0 - 1)] - 1}{4}$$

Answer 3

Generally, if the sequence $\{a_n\}$ is given by $$a_{n+1}=\frac{Ca_n+D}{Aa_n+B}.$$ Let $$f(x)=\frac{Cx+D}{Ax+B}$$ be an rational function of degree $1$, then we have the following conclusion:

$(1)$ If $f(x)$ has only one fixed point $x_0$, then $$\left\{\frac{1}{a_{n}-x_0}\right\}$$ forms a tolerance of $\frac{-A}{Ax_0-C}$ arithmetic progression.

$(2)$ If $f(x)$ has two fixed points $x_1,x_2$, then $$\left\{\frac{a_{n}-x_1}{a_{n}-x_2}\right\}$$ becomes a common ratio of $\frac{Ax_1-C}{Ax_2-C}$ geometric progression.

You can check that $$\frac{1}{a_{n+1}-\frac{1}{2}}-\frac{1}{a_{n}-\frac{1}{2}}=4.$$ That is to say, $\left\{\frac{1}{a_{n}-\frac{1}{2}}\right\}$ forms a tolerance of $4$ arithmetic progression. So $$\frac{1}{a_{n}-\frac{1}{2}}=\frac{1}{a_{1}-\frac{1}{2}}+4(n-1),\ n\geq 1 .$$ Easy computation gives that $$a_n=\frac{(2n-1)a_1-(n-1)}{(4n-4)a_1-(2n-3)},\ n\geq 1.$$