Questions involving the inductive sequence, $x_1 = 7$, $x_{n+1} = \sqrt{3x_n + 4}$ for $n \geq 1$

Question

I'm a university student and I've got a test coming up and revising. I've answered my question as best I could but if anyone has any tips to set me on the right course that would be fantastic! My logic will likley be a little messy so I apologise in advance.

The sequence in question is...

$$x_1 = 7, x_{n+1} = \sqrt{3x_n + 4}, n \geq 1$$

The question asks me to show that $4 < x_n < 10$. I'll try solve inductivly. We establish the base case to be $x_1 = 7$ we assume $4<x_n<10$ therefore for case $x_n \rightarrow 10$...

$$ x_{n+1} = \sqrt{30+4}= \sqrt{34}, 4<\sqrt{34} < 10$$

And for the case $x_n \rightarrow 4$...

$$ x_{n+1} = \sqrt{16}= 4, 4<\sqrt{16} < 10$$

I am confused here, the range would seem make sense if $4 \leq x_n <10$. Moving on, the question now tells me to prove $x_n$ is decreasing.

$$x_{n+1} - x_n = \sqrt{3x_n + 4} - x_n $$

Here I take the highest number in the range (10) and the lowest (4) and observe $x_n=4$ yields 0 and $x_n = 10$ yeilds a negative result therefore it must be decreasing. We can then deduce by the monotone convergence theorem that $\{ x_n \}$ converges, so $x_n \rightarrow x$ and $x_{n+1} \rightarrow x $. The uniqueness of limits implies...

$x = \sqrt{3x+4}$

Which has root, $x=4$, and is thus the limit.

Answer 1

Induction step:

If $4<x_k < 10$, then we have $12+4 < 3x_k+4 < 30+4<36$, hence by taking square root we have

$$4<x_{k+1}<6$$

which implies that $$4 < x_{k+1}< 10.$$

Remark: You should not be using $\to$ in your proof, we are just taking a single step.

To prove decreasing:

$$x_{n+1}-x_n = \sqrt{3x_n+4}-x_n=\frac{3x_n+4-x_n^2}{\sqrt{3x_n+4}+x_n}=\frac{(-x_n-1)(x_n-4)}{\sqrt{3x_n+4}+x_n}\le 0$$

Remark: We have to do more besides checking the end points.

Answer 2

Hint. Note that if $x_n>4$ then $$x_{n+1} - x_n = \sqrt{3x_n + 4} - x_n=\frac{3x_n + 4 - x^2_n}{\sqrt{3x_n + 4} + x_n}=-\frac{(x_n-4)(x_n+1)}{\sqrt{3x_n + 4} + x_n}<0$$ and therefore $(x_n)_n$ is strictly decreasing.

Answer 3

Decreasing:

$f(x)=(3x+4)^{1/2}; $

$f'(x)=(1/2)(3x+4)^{-1/2} >0$, increasing .

$ x_2=5; x_1=7;$

$x_2<x_1;$

$x_3 =f(x_2) < f(x_1)= x_2;$

$x_4 =f(x_3) < f(x_2)=x_3;$

Define inductively:

$x_{n+1}< x_n$;

$x_{n+2}=f(x_{n+1})<f(x_n) =x_{n+1}.$