$a_1=\sqrt{6}$ , $a_{n+1} = \sqrt{6+a_n}$

A sequence $(a_n)$ is defined by $a_n > 0$ and $a_{n+1} = \sqrt{6+a_n}$ for $n\ge1$. Show that

$1)$ the sequence $(a_n)$ is monotone increasing if $0 < a_1 < 3$

$2)$ the sequence $(a_n)$ is monotone decreasing if $a_1 > 3$

I know how to verify it by substituting values for $n$ but to prove generally, I don't know. I know that the limit of the sequence is $3$ which is obtained by the positive root of $x^2 - x -6 =0$.

We usually use $a_{n+1} - a_n$ to check if a sequence is monotone increasing our monotone decreasing. But here, since intervals are involved, I don't know how to proceed.

Thank you in advance.

Solution 1:

It is easy to show by induction that in case 1) $0<a_n<3$ for all $n$ and in case 2) $a_n>3$ for all $n$. Now $$ a_{n + 1} - a_n = \sqrt {6 + a_n } - a_n = \frac{{6 + a_n - a_n^2 }}{{\sqrt {6 + a_n } + a_n }} = \frac{{(3 - a_n )(a_n + 2)}}{{\sqrt {6 + a_n } + a_n }}. $$ Taking into account my first sentence, the monotonicity properties follow readily.

Solution 2:

Let

$$f(x):=\sqrt{6+x}$$

The proof by @dxiv is perfect with his key expression that I write under the following form:

$$\dfrac{a_{n+1}−a_n}{a_n-a_{n-1}}=\dfrac{1}{\sqrt{6+a_n}+\sqrt{6+a_{n-1}}} \color{red}{\approx \dfrac{1}{2\sqrt{6+L}} = f'(L)}$$

with added terms in red, providing an approximation for large enough values of $n$. In this way, one (re)discovers the fact that the rate of convergence of an iterative sequence $a_{n+1}=f(a_n)$ is governed by the derivative of $f$ at its limit value, here $L=3$.

I would like here to give a complementary graphical "proof without words" (in fact I have added a lot of words!).

Fig. 1: The two curves with equations $y=f(x)$ and $y=x$ intersect in fixed point $(L,L)=(3,3)$ with two staircase patterns: ascending in red (with $a_0=0 < L$), descending in blue (with $a_0=6 > L$), both converging to fixed point.

where the upper image is a zoom on the lower image. Fixed point iteration

$$a_{n+1}=f(a_n)$$

is materialized by staircase patterns, each step joining points which are alternately on the curve and on the bissector line:

$$\underbrace{(a_n,f(a_n)=a_{n+1})}_{\text{on the curve}} \ \ \to \ \ \underbrace{(a_{n+1},a_{n+1})}_{\text{on the biss. line}} \ \ \to \ \ \underbrace{(a_{n+1},f(a_{n+1})=a_{n+2})}_{\text{on the curve}} \ \ \to \ \ \text{etc.}$$

If one starts below (resp. above) limit point $3$, we will have an ascending (resp. descending) staircase.

Remark: See this answer of mine to this question in a similar (but very particular) issue.

Solution 3:

If $a_n > 3$ then $$a_n = \sqrt{a_n \cdot a_n} > \sqrt{3 a_n} = \sqrt{2 \cdot a_n + a_n} > \underbrace{\sqrt{6 + a_n}}_{= a_{n+1}} > \sqrt{6 + 3} = \sqrt{9} =3 $$ and if $0 < a_n < 3$ then $$a_n = \sqrt{a_n \cdot a_n} < \sqrt{3 a_n} = \sqrt{2 \cdot a_n + a_n} < \underbrace{\sqrt{6 + a_n}}_{= a_{n+1}} < \sqrt{6 + 3} = \sqrt{9} =3 $$ by strict monotony of the square root (in each non-trivial step an $a_n$ is replaced by a $3$). The desired statement follows by complete induction on $n$.