How to prove the sequence given by $a_{n+1}=s+a_n^2$ is monotonic increasing?

Question

Let $s$ be $0\:\le \:s\le \:\frac{1}{4}$ and consider this sequence:
$a_1\:=\:s$
$a_{n+1}\:=\:s\:+\:a_n^2$

I want to prove that is monotonic sequence, so I thought about induction or assume in negative and get a contradiction but I always get stuck. can someone guide me for the method how to prove this? tnx!

Answer 1

Induction:

$$a_2:=s+a_1^2\ge a_1\iff a_1^2-a_1+s\ge 0\stackrel{a_1=s\ge 0}\iff \Delta:=1-4s\le 0\;$$

and the last inequality is true since $\;s\le\frac14\;$ .

Now

$$a_{n+1}:=s+a_n^2\ge a_n\iff a_n^2-a_n+s\ge 0\;\stackrel{\text{ind. Hyp.}\implies a_n\ge 0}\iff\;\Delta:=1-4s\ge 0$$

and we're done.

Answer 2

You can use induction to prove that $0 \le a_n \le a_{n+1}$ for all $n$. The base case holds because $a_1 = s \le s + a_1^2 = a_2$ and $s \ge 0$.

Now assume that $0 \le a_n \le a_{n+1}$. Square both sides to get $0 \le a_n^2 \le a_{n+1}^2$, and add $s$ to both sides to conclude $0 \le a_{n+1} \le a_{n+2}$.

Answer 3

Proof by induction:

base case: $n=1$ $a_1=s, a_2=s+a_1^2 = s + s^2 \geq s = a_1$

$n \implies n+1$: $a_{n+1} = s + a_n^2 \geq 2 \sqrt{sa_n^2} = 2 a_n \sqrt{s} \geq a_n 2 \sqrt{\frac{1}{4}} = a_n$, as needed.

Answer 4

Let $f(x) = s+x^2$. Note that if $ x \ge 0$ then $f(x) \ge 0$. Also, $f'(x) = 2x$, so $f$ is non-decreasing on $[0, \infty)$. In particular, if $0 \le x\le y$ then $0 \le f(x) \le f(y)$.

We have $0 \le s = a_1 \le a_2 = f(a_1) = s^2+s$.

Now suppose $0 \le a_n \le a_{n+1} = f(a_n)$. Then $0 \le a_{n+1}=f(a_n) \le f(f(a_n)) = f(a_{n+1}) = a_{n+2}$.

Hence $0 \le a_n \le a_{n+1}$ for all $n$.

Note that this result is true for any $s \ge 0$.