Explicit formula of $a_{n+1}=\frac{3a_n+3}{a_n+3}$, $a_1 = 3$

Question

The original question is the following:

Suppose that $a_1 = 3$ and that $a_{n+1} = \frac{3a_n+3}{a_n+3}$. By considering $\frac{a_{n+1} - \sqrt{3}}{a_{n+1} + \sqrt{3}}$, find a formula for $a_n$ in terms of $a_1$ and $n$, and deduce that $a_n \rightarrow \sqrt{3}$.

(Source: Alan F. Beardon, "Limits - A New Approach to Real Analysis", 1997, Springer)

Here is my attempt:

Let $b_n = \frac{a_{n+1} - \sqrt{3}}{a_{n+1} + \sqrt{3}}$.

$b_2 = \frac{3-\sqrt{3}}{3+\sqrt{3}} = 2 - \sqrt{3}$

$b_3 = \frac{2-\sqrt{3}}{2+\sqrt{3}} = 7 - 4\sqrt{3}$

$b_4 = \frac{\frac{9}{5} - \sqrt{3}}{\frac{9}{5} + \sqrt{3}} = \frac{\frac{156}{25} - \frac{18}{5} \sqrt{3}}{0.24} = 26 - 15\sqrt{3}$

$b_5 = \frac{\frac{7}{4} - \sqrt{3}}{\frac{7}{4} + \sqrt{3}} = \frac{6.0625 - 3.5 \sqrt{3}}{0.0625} = 97 - 56 \sqrt{3}$

Let $b_n = c_n - \sqrt{3} d_n$

$7 = 2\times 4 - 1 \\ 26 = 7 \times 4 - 2 \\ 97 = 26 \times 4 - 7$

$\therefore c_1 = 1$, $c_2 = 2$. For $n \geq 1$, $c_{n+2} = 4c_{n+1} - c_n$

Similarly, $4 = 1 \times 4 - 0 \\ 15 = 4 \times 4 - 1 \\ 56 = 15\times 4 - 4$

$\therefore d_1 = 0$, $d_2 = 1$. For $n \geq 1$, $d_{n+2} = 4d_{n+1} - d_n$

$\begin{split} c_n & = 4 (c_{n-2} - c_{n-3}) - c_{n-2} \\ & = [4^2 - 1] c_{n-2} - 4 c_{n-3} \\ & = [4^3 - 2(4)] c_{n-3} - [4^2 - 1] c_{n-4} \\ & = [4^3 - 3(4)^2 + 1]c_{n-4} - [4^3 - 2(4)] c_{n-5} \\ & = [4^5 - 4(4)^3 + 3(4)]c_{n-5} - [4^4 - 3(4)^2 + 1] c_{n-6} \end{split}$

However, I can't find for the pattern between the coefficients of sequences. Any hint for the next steps? Thanks.

Answer 1

You don't use the hint in a useful way.

$$\frac{a_{n+1}-\sqrt 3}{a_{n+1}+\sqrt3}=\frac{\dfrac{3a_n+3}{a_n+3}-\sqrt 3}{\dfrac{3a_n+3}{a_n+3}+\sqrt3}=\frac{(3-\sqrt3)a_n+3-3\sqrt3}{(3+\sqrt3)a_n+3+3\sqrt3}=\frac{3-\sqrt3}{3+\sqrt3}\frac{a_n-\sqrt3}{a_n+\sqrt3}.$$

Hence by induction,

$$\frac{a_{n+1}-\sqrt 3}{a_{n+1}+\sqrt3}=\left(\frac{3-\sqrt3}{3+\sqrt3}\right)^{n+1}.$$

Answer 2

Further answering the question:

$$\displaystyle\lim_{n \rightarrow \infty} \dfrac{a_n - \sqrt{3}}{a_n + \sqrt{3}} = \displaystyle\lim_{n \rightarrow \infty} \left( \dfrac{3-\sqrt{3}}{3+\sqrt{3}}\right)^n = 0$$ $$\displaystyle\lim_{n \rightarrow \infty} \left( 1 - \dfrac{2 \sqrt{3}}{a_n + \sqrt{3}} \right) = 0$$ $$2\sqrt{3} \displaystyle\lim_{n \rightarrow \infty} \dfrac{1}{a_n + \sqrt{3}} = 1$$ $$\displaystyle\lim_{n \rightarrow \infty} \dfrac{1}{a_n + \sqrt{3}} = \dfrac{1}{2\sqrt{3}}$$ $$\displaystyle\lim_{n \rightarrow \infty} \left( a_n + \sqrt{3} \right) = 2\sqrt{3}$$ $$\displaystyle\lim_{n \rightarrow \infty} a_n = \sqrt{3} $$