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 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,390 Thanks: 100 
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: 7,329 Thanks: 2452 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: 1,956 Thanks: 547  Quote:
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 $\displaystyle xε$ and finite terms of $\displaystyle a_n$ that are greater than $\displaystyle x+ε$ ?

May 31st, 2017, 08:13 AM  #7  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,390 Thanks: 100  Quote:
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: $\displaystyle xa_{n}< \epsilon$ 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: 1,956 Thanks: 547  Quote:
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 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,390 Thanks: 100  

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inf, lim, sequence 
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