User Name Remember Me? Password

 Calculus Calculus Math Forum

 May 30th, 2017, 09:52 AM #1 Member   Joined: Apr 2014 From: Greece Posts: 58 Thanks: 0 lim sup and lim inf of a sequence I 'm confused with the superior and inferior limit of a sequence. If I get it right, the $\displaystyle lim supa_{n}$ is the supremum of the set of convergence points of every possible subsequence of $\displaystyle a_{n}$ right? Does that mean that $\displaystyle infa_{n}\leq liminfa_{n}\leq lim supa_{n}\leq supa_{n}$ is true in any case? If $\displaystyle a_{n}$ diverges then can the above statement be wrong? (meaning that some subsequnce of $\displaystyle a_{n}$ converges to a point out of $\displaystyle a_{n}$ boundaries) Sorry if my questions are stupid I'm just super confused for some reason. May 30th, 2017, 10:54 AM #2 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 You are right. Frankly, I get confused myself. That's why I hate jargon. I prefer least upper bound and greatest lower bound, and upper and lower limits. May 30th, 2017, 11:48 AM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,620 Thanks: 2609 Math Focus: Mainly analysis and algebra As I understand it, $\displaystyle \sup_{k \ge n} a_k = \sup {a_k: k \ge n}$ is the Least Upper Bound of the (sub-)sequence $(a_n,a_{n+1},a_{n+2},a_{n+3},\ldots)$. Then $$\limsup a_n = \lim_{n \to \infty} \left( \sup_{k \ge n} a_k \right)$$ That is, if we create a sequence $\{b_n\}$ where each $b_n$ is the supremum of all the $a_k$ from $a_n$ onwards, then $\displaystyle \limsup a_n = \lim_{n \to \infty} b_n$ (where the limit exists). Thus for the sequence $a_n = \frac1n$, we have $\displaystyle b_n = \sup_{k \ge n} a_k = \frac1n$ and thus $\limsup a_n = 0$ Your inequality is always true. May 30th, 2017, 03:49 PM   #4
Senior Member

Joined: Aug 2012

Posts: 2,205
Thanks: 647

Quote:
 Originally Posted by Vaki I 'm confused with the superior and inferior limit of a sequence. If I get it right, the $\displaystyle lim supa_{n}$ is the supremum of the set of convergence points of every possible subsequence of $\displaystyle a_{n}$ right?
No, it's the limit of the sequence of sups of the tails of the sequence.

If you allowed every possible subsequence, then because some subsequence contains the sup of the original sequence, the tail sup would never change.

Also you can mark up lim and sup as keywords so it comes out $\lim \sup a_n$. You can Quote my post to see the markup. May 31st, 2017, 04:45 AM #5 Member   Joined: Apr 2014 From: Greece Posts: 58 Thanks: 0 So basically are we just looking for all the relative maximum and minimumm of the sequence $\displaystyle a_{n}$ and make a new sequence for the maxima and one for the minima out of these terms? And the limit of these new sequence as n goes to infinity is the $\displaystyle \lim \sup a_n$ and $\displaystyle \lim \inf a_n$ If that's true then $\displaystyle \lim \sup a_n \leq \sup an_{n}$ in any case. And also if $\displaystyle \lim \sup a_n=\lim \inf a_n$ then the sequence converges right? May 31st, 2017, 04:59 AM #6 Member   Joined: Apr 2014 From: Greece Posts: 58 Thanks: 0 Let me ask you one more thing. If we have a bounded sequence $\displaystyle a_n$ and $\displaystyle x=\lim \sup a_n$ How can we prove that for $\displaystyle ε>0$ there are infinite terms of $\displaystyle a_n$ that are greater than $\displaystyle x-ε$ and finite terms of $\displaystyle a_n$ that are greater than $\displaystyle x+ε$ ? May 31st, 2017, 08:13 AM   #7
Banned Camp

Joined: Mar 2015
From: New Jersey

Posts: 1,720
Thanks: 124

Quote:
 Originally Posted by Vaki Let me ask you one more thing. If we have a bounded sequence $\displaystyle a_n$ and $\displaystyle x=\lim \sup a_n$ How can we prove that for $\displaystyle ε>0$ there are infinite terms of $\displaystyle a_n$ that are greater than $\displaystyle x-ε$ and finite terms of $\displaystyle a_n$ that are greater than $\displaystyle x+ε$ ?
An infinite bounded sequence can have one or more limit points.The largest of these is the upper limit and the least of these is the lower limit; the sequence also has a least upper bound and a greatest lower bound.

From these definitions it is obvious that:
$\displaystyle GLB \leq LL \leq UL \leq LUB$

If $\displaystyle x$ upper limit of $\displaystyle a_{n}$, then:
$\displaystyle |x-a_{n}|< \epsilon$ for an infinte number of $\displaystyle a_{n}$ by definition of limit point.

If there were an infinite number of $\displaystyle a_{n}$ greater than $\displaystyle x+\epsilon$, x would not be an UL. May 31st, 2017, 10:29 AM   #8
Senior Member

Joined: Aug 2012

Posts: 2,205
Thanks: 647

Quote:
 Originally Posted by zylo An infinite bounded sequence can have one or more limit points.The largest of these is the upper limit and the least of these is the lower limit; the sequence also has a least upper bound and a greatest lower bound.
If the sequence is 50, -50, 1, 1/2, 1/4, 1/8/ 1/16, ...

what is the least upper bound? The greatest lower bound? The upper limit? The lower limit? The limit? May 31st, 2017, 10:44 AM   #9
Banned Camp

Joined: Mar 2015
From: New Jersey

Posts: 1,720
Thanks: 124

Quote:
 Originally Posted by maschke if the sequence is 50, -50, 1, 1/2, 1/4, 1/8/ 1/16, ... What is the least upper bound? The greatest lower bound? The upper limit? The lower limit? The limit?
50, -50, 0, 0, 0 Tags inf, lim, sequence Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Tomedb Algebra 1 November 3rd, 2016 03:17 PM Rahul k New Users 2 May 12th, 2015 05:51 PM Algebra 3 April 21st, 2012 07:32 PM elim Real Analysis 8 May 23rd, 2010 12:23 PM babyRudin Real Analysis 6 October 10th, 2008 11:11 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top       