My Math Forum upper limit = lower limit implies convergence

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 May 29th, 2017, 11:23 AM #1 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,082 Thanks: 87 upper limit = lower limit implies convergence If $\displaystyle \overline \lim_{n\rightarrow \infty}S_{n}=\underline \lim_{n\rightarrow \infty}S_{n}=S$ the sequence converges. It seems obvious, but does anyone have a formal proof? Definition from: Upper Limit -- from Wolfram MathWorld "An upper limit of a series upper $\displaystyle \lim_{n\rightarrow \infty} S_{n}=\overline {\lim_{n\rightarrow \infty}}S_{n}=k$ is said to exist if, for every $\displaystyle \epsilon>0, |S_{n}-k|<\epsilon$ for infinitely many values of n and if no number larger than k has this property." Similarly for lower limit.
 May 29th, 2017, 02:36 PM #2 Senior Member   Joined: Aug 2012 Posts: 1,434 Thanks: 353
 May 29th, 2017, 06:09 PM #3 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,082 Thanks: 87 The referenced article is unintelligible at first reading. However, it states that $\displaystyle \inf_{n>N}\leq a_{n} \leq \sup_{n>N}$ But inf is greatest lower bound which is not the same as lower limit. See OP. So the article is irrelevant. I note Rudin mentions the OP theorem but doesn't prove it.
 May 29th, 2017, 06:23 PM #4 Senior Member   Joined: Aug 2012 Posts: 1,434 Thanks: 353 I would love to hack this one out but haven't got the time at the moment. I thought my link was about lim inf and lim sup. Those are alternate names for what you're calling lower and upper limits respectively. I didn't read the link very closely. You can probably google around and find a proof. If infinitely many values are within epsilon of the lower limit and also the upper limit then the sequence can't converge. If I had to do this I'd start there and bang it out from the defs. ps I read the link. There are a couple of good posts showing approaches to this problem. It's always a bit of work to do lim inf and lim sup proofs, the notation and concepts are a little slippery. Last edited by Maschke; May 29th, 2017 at 06:26 PM.
 May 29th, 2017, 07:26 PM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,877 Thanks: 2240 Math Focus: Mainly analysis and algebra Wolfram is clear that the Upper Limit is not the same as the Supremum Limit, but that is not the same as saying that it doesn't apply because $\inf a_n$ is not the lower limit. The key point about the Upper/Lower Limits definitions is the same as the Supremum/Infimum Limits: they define a region within which the sequence ultimately oscillates. So the proof is going to start with pointing out that since both Upper and Lower Limits exist, the sequence doesn't diverge to $\pm\infty$. Given that, I think we can then go for the contra-positive and show that a divergent sequence $a_n$ must have $$\overline{\lim_{n \to \infty}} a_n \ne \underline{\lim_{n \to \infty}} a_n$$
 May 29th, 2017, 09:20 PM #6 Senior Member   Joined: Aug 2012 Posts: 1,434 Thanks: 353 Here's another SE thread with some good insights. I didn't read much of it, but it looks clearer. Lim inf and lim sup are notoriously slippery concepts. You have to work at this problem a little, you can't just see a proof and go "Oh, I got it," without a fair amount of work. https://math.stackexchange.com/quest...rgent-sequence ps ... Another clue from that thread. I have the visualization now. If $(a_n)$ is bounded then it has an inf and a sup. We can visualize these as two horizontal lines in the plane, between which all of the terms of $(a_n)$ live. It's like the outer envelope of all the points. Now define $u_n = \sup \{a_k : k \geq n\}$. In other words $u_n$ is the "sup of the $n$-th tail." Likewise $d_n$ is defined as the inf of the $n$-th tail. Now we define $\lim \sup a_n = \lim_{n \to \infty} u_n$ and likewise for lim inf. So lim sup is the "limit of the tail sups." and lim inf is the limit of the tail infs. Why do these limits exist? Needs proof, but the tail sups are monotonic and so are the tail infs and that leads to the proof. Now we can imagine the $u_n$'s and $d_n$'s as creating a shrinking envelope. If we connected the dots, we'd have a funnel narrowing down as $n$ increases. If the gap between $u_n$ and $d_n$ is greater than zero, the sequence doesn't converge; and if the gap goes to zero, the sequence converges. This is now obvious intuitively and you could bang out a formal proof if you were so inclined. Last edited by skipjack; May 29th, 2017 at 10:26 PM.
 May 29th, 2017, 10:22 PM #7 Senior Member   Joined: Aug 2012 Posts: 1,434 Thanks: 353 My edit window timed out before I could fix the markup. The broken line is the definition of lim sup as the limit of the tail sups: $\lim \sup a_n = \lim_{n \to \infty} u_n$ I hope it's clear what's going on. The tail sups are monotone decreasing and the tail infs are monotone increasing. If their difference goes to zero, they squeeze all but finitely many elements of the sequence into any epsilon-ball. All of this is not the proof, just the visualization. You still have to nail down the symbology. Last edited by Maschke; May 29th, 2017 at 10:24 PM.
May 30th, 2017, 09:57 AM   #8
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Your upper and lower lines are limit points for infinite numbers of s_{n}. but you need a third or more lines to indicate other possible limit points if upper limit is unequal to lower limit.

The existence of inf and sup doesn't rule out limit points between them.

Assume inf=sup. Then there is no limit point between them and $\displaystyle |s_{n}-s|<\epsilon$ for n>N? No. All you can say is for an infinite number of n greater than N, not for ALL n>N.

EDIT:
Quote:
 Originally Posted by Maschke If $(a_n)$ is bounded then it has an inf and a sup. We can visualize these as two horizontal lines in the plane, between which all of the terms of $(a_n)$ live. It's like the outer envelope of all the points.
That's the definition of greatest lower bound and least upper bound.

Last edited by zylo; May 30th, 2017 at 10:04 AM.

 May 30th, 2017, 09:59 AM #9 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,877 Thanks: 2240 Math Focus: Mainly analysis and algebra That why I'd go for contra-positive. All those problems go away I think.
May 30th, 2017, 10:12 AM   #10
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Quote:
 Originally Posted by v8archie That why I'd go for contra-positive. All those problems go away I think.
OK. So you assume upper and lower limits are equal but the sequence doesn't converge. Now what?

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