Determining convergence for sequence $a_1 = 3, \ a_{n+1} = \sqrt{3a_n - 2}$

Question

Problem

Given the sequence where $a_1 = 3$ and $a_{n+1} = \sqrt{3a_n -2}$, show that it converges, and find the limit $\lim\limits_{n\to\infty}a_n$.

Progress

The way we are taught to solve these, is to say that if we let $x=\lim\limits_{n \to \infty} a_n $

$$\lim\limits_{n \to \infty} a_{n+1} = \sqrt{3x-2} = x$$

since $a_{n+1} = a_n$ as $n\to\infty$.

Solving this equation, we get that $x=1 \vee x=2$, but from there I have to figure out which.

A quick Python-script seems to show that it progresses asymptotically towards 2, and no lower. This is no proof, but a basis for a hypothesis at least.

Induction?

My next idea is a proof by induction, which summarizes as:

Hypothesis: $n \geq 1 \Rightarrow a_n > 2$. Holds for $n=1, 2$ by trial.

Then I want to show that $a_n > 2 \Rightarrow a_{n+1} > 2$, which is to say that if $\sqrt{3a_{n-1} - 2} > 2$ then $\sqrt{3a_n - 2} > 2$.

For the latter inequality, we can expand: $\sqrt{3a_n - 2} = \sqrt{3\sqrt{3a_{n-1}-2}-2} = \sqrt{9a_{n-1} - 8} = \sqrt{3a_{n-1}-2}\cdot \sqrt{\frac{9a_{n-1} - 8}{3a_{n-1}-2}} > 2$

Here is where I need some validation. In this final product, the first factor is greater than 2 by the initial trials, and the second factor is greater than 1 for all $a_{n-1} > 1$ (since it has a root at 8/9, and is strictly increasing).

Question

Does this proof hold? And are there other, simpler ways to conclude the same? Or was the hypothesis wrong in the first place?

Answer 1

$a_1=3$ and $a_2=\sqrt 7 \leq a_1$, assume if $a_{k+1} \leq a_k$ then

$$3a_{k+1} \leq 3a_k \implies 3a_{k+1} -2 \leq 3a_k -2 \implies \sqrt{(3a_{k+1} -2)} \leq \sqrt {(3a_k -2)} \implies a_{k+2} \leq a_{k+1} $$ so by induction, $a_n$ is decreasing.

if $a_1=3 \geq 2$ assume $a_k \geq 2 \implies 3a_k-2 \geq 4 \implies a_{k+1} \geq 2.$ So $a_n$ is monotonically decreasing and bounded below so by the Monotone convergent theorem it is convergent.

Let $\lim n \to \infty (a_{n+1}) = l = \lim n \to \infty (a_{n}) \implies l=2.$

Answer 2

The question can be put in a general scheme, resembling the Banach fixed point theorem. This approach implies the exponential rate of the convergence of the sequence to its limit.

Let $a_{n+1}=f(a_n),$ where $0<f'(x)\le r<1$ for $c<x<d.$ Assume $f(x_0)=x_0$ for some $c<x_0<d.$ If $c<a_1<d$ then $c<a_n<d$ and $a_n$ is convergent to $x_0.$

Indeed, there are two cases $a_1\in (x_0,d)$ or $a_1\in (c,x_0].$ We will consider the first case, the second can be dealt with similarly. We will prove by induction that $a_n\in (x_0,d).$ Assume $a_n\in (x_0,d).$ We have $$a_{n+1}-x_0=f(a_n)-f(x_0)\\ =f'(\xi_n)\, (a_n-x_0)\le r(a_n-x_0)\quad (*)$$ As $f'>0$ we get $a_{n+1}> x_0.$ Moreover $$a_{n+1}\le ra_n+(1-r)x_0<d$$ This concludes the induction step.

The inequality $(*)$ implies $$0<a_{n+1}-x_0\le r^n( a_1-x_0)$$ Thus $\lim a_n=x_0.$

In the OP case $f(x)=\sqrt{3x-2}$, $c={3\over 2}$ , $d=\infty$ and $x_0=2.$ Moreover $$0<f'(x)={3\over 2\sqrt{3x-2}}<{3\over 2\sqrt{2.5}}=:r<1$$