$a_{n+1}=\frac1{a_n}+\frac{a_n}2,$ prove that the sequence is bounded.

Question

Let $$a_1=2\quad\text{and}\quad a_{n+1}=\frac1{a_n}+\frac{a_n}2\quad(\forall n\ge1).$$

i)Show that the sequence (a_n) is bounded

ii) Show that (a_n) is decreasing

What i've tried: i) Most likely going to be via induction? I've tried setting my claim as 1<a_n<=2 , but can't seem to get anywhere with it.

ii) a_n+1-a_n = 1/a_n -a_n/2 . Now need to show that this is <0 but can't see a way to do this without having bounds from (i)

thanks for any help

Does this answer your question? Real infinite sequence - found using an Approach0 search. In particular, the currently accepted answer there shows the sequence bounded below by $\sqrt{2}$ and is non-increasing. Note there are other basically duplicate questions found, e.g., ... — John Omielan, Jan 01 '23 at 22:41
(cont.) Prove that $\lim_{n \rightarrow +\infty} v_{n+1} = \sqrt{2}$? and Proving the limit of a recursive sequence. Also, there're questions with a starting value of $a_1=1$ instead at How to prove the sequence $x_{n+1} = \frac{x_n}2 + \frac 1{x_n}$ is a Cauchy sequence & Prove that ${x_1=1,,x_{n+1}=\frac {x_n}2+\frac 1{x_n} }$ converges when $n \to \infty$. Finally, there's the related AoPS thread of for which values this succession converges? (re: valid starting values). — John Omielan, Jan 01 '23 at 22:49
It helps a lot, however currently i only have a lower bound for a_n, i need to show its bounded above and below. With the lower bound part ii comes immediately so it's just finding this last bit — Skark123, Jan 01 '23 at 23:01
Welcome to Math SE. I'm glad you found those other posts to be helpful. Also, note proving the sequence is non-increasing automatically means it's bounded above by $a_1=2$. — John Omielan, Jan 01 '23 at 23:04
Yes you're right, not sure how i didn't spot that. Thanks for the help — Skark123, Jan 01 '23 at 23:14

score 0 · Answer 1 · answered Jan 01 '23 at 23:51

It is clear that $a_n>0$ implies $a_{n+1}>0$. Since $a_1=2>0$, $a_n>0,\forall n\in\mathbb{N}$.

To prove $a_n$ is decreasing, observe $$a_{n+1}-a_n=\frac{1}{a_n}-\frac{a_n}{2}<0 \iff a_n^2>2.$$

We study when $a_{n+1}^2>2$: $$a_{n+1}^2=\frac{1}{a_n^2}+\frac{a_n^2}{4}+1>2\iff 4+a_n^4>4a_n^2\iff (a_n^2-2)^2>0\iff a_n^2\ne2$$

$a_1^2=4>2$, hence $a_n^2>2$, $\forall n\in\mathbb N$, so $a_n$ is strictly decreasing. It follows that $a_n\le a_1=2$, so $0<a_n\le2$ is bounded.

score 0 · Answer 2 · answered Jan 02 '23 at 00:35

This is a nonstandard approach, but I include it as a curiosity:

A closed form representation for your function can be written in the form $x \mapsto \sqrt{2} \coth(m_xc)$ where $c \in \mathbb{R}^+$ is a constant and $m_x$ is increasing as $x$ increases. Since the hyperbolic cotangent is a decreasing function for $x>0$ and bounded below by $1$, parts (ii) and (i) of your question both follow. In particular, your terms are strictly decreasing and bounded below by $\sqrt{2}$.

Here is a graph in Desmos:

$a_{n+1}=\frac1{a_n}+\frac{a_n}2,$ prove that the sequence is bounded.

2 Answers2