Use induction to demonstrate properties of the sequence $a_1 = 4$, $a_{n+1} = \sqrt{a_n+6}$

Question

I just can't understand how to approach this question, which is probably very easy. Tried to look online for solutions but didn't come across something helpful. Here's the question:

There is a series where: $a_1 = 4$, $a_{n+1} = \sqrt{a_{n}+6}$.

Prove by induction:

1. For every natural $n$ number: $a_{n}>3$

2. Prove that $a_{n} \geq a_{n+1}$ for every natural n

I understand the idea of induction, as finding something valid for the first member of the group, assuming it's true for the $n$ object and showing it's indeed correct for the $(n+1)$-st object. I just can't understand how to pull this off on this question.

Answer 1

For the first part, use $x>3\implies\sqrt{x+6}>\sqrt{3+6}=3$ in the inductive step. For the second, you don't need an extra proof by induction: use the result of the first part together with$$x>3\implies x^2-x-6=(x-3)(x+2)>0\implies x+6<x^2\implies\sqrt{x+6}<x.$$

Answer 2

Base cases are trivial then for the induction step for point $1$ assume that $a_n >3$ then

$$a_{n+1} = \sqrt{a_{n}+6} >\sqrt{3+6}=3$$

and for point $2$ assume $a_{n} \geq a_{n+1}$ then

$$a_{n+2} = \sqrt{a_{n+1}+6}\le \sqrt{a_{n}+6}=a_{n+1}$$

Answer 3

Hints for the inductive steps:

1. If $a_n>3$ then $a_n+6>9$, so $a_{n+1}^2>9$, so $a_{n+1}>3.$

2. If $a_n>3$ then $(a_n-3)(a_n+2)>0$, so $a_n^2-a_n-6>0$, so $a_n>\sqrt{a_n+6}=a_{n+1}$.

Answer 4

Supoose $a_n\geq a_{n-1}$. Then, for $a_{n+1}\geq a_{n}$, we have $\sqrt{a_n+6}\geq\sqrt{a_{n-1}+6}$ which completes the proof, because we assumed $a_n\geq a_{n-1}$.

Suppose $a_n\geq 3$ Then $a_{n+1}\geq3$ is equivalent to $\sqrt{a_n+6}\geq3$ and by squaring the proof is done.