# Thread: Limit Sup

1. ## Limit Sup

According to my book, the limit sup s_n is always less then or equal to the limit s_n

I don't understand this. If it is the sup, i expect that it is always GREATER than or equal to limit s_n. Furthermore, what bothers me is that the sequence V_N decreases (or is equal to it's previous term), where V_N = sup{ s_n : where n > N }. BUT WHY? Should the V_N not be all the same? because it doesn't matter if n> 20 or n > 100 or n > 100000, the sup is the least upper bound, so whatever the top is for n > 20, should be the same for n > 100000 because 10000 > 20 anyway. In the case of the sequence U_N for the inf, i can understand why U_N increases as N increases because your interval changes. But how does the interval change for V_N ?? The upper bound for n > 20 seems intuitively to be the same for the upper bound where n > 21 say. PLease help.

2. For $V_n$ consider $s_1=2$ and $s_n=\frac{n}{n+1}$ for $n>2$. In general removing terms will always decrease the limsup since you might have removed a term that was equal to the limsup.

Also if $\lim s_n$ exist, it must be equal to both $\limsup s_n$ and $\liminf s_n$. So I don't see how they say that the limsup (which always exist) is less than or equal to the limit (which might not even exist).

3. im sorry, i made a mistake. the book says that limit sup is less than or equal to sup{s_n : n is a member of the naturals}. i interpreted the second part as being limit s_n. but i see it now.

However, i still don't see why V_n is a decreasing sequence. V_2 should be the same as V_100 ? if we want the least upper bound, how do we know where the upper bound is if it's not at infinity (thus being the same) ? n > 2 only specify where to start, and is not concerned with the upper bound. i'm confused.

4. V_n is the sup NOT the limsup.

5. No one seem to be seeing where the point of my confusion lies. Suppose we take the sequence 1/n for n bigger than 0. s_1 is bigger then s_2 which is bigger then s_3 and so on. In other words, it is a nonincreasing monotone sequence. By def, U_N = inf{s_n : n > N} . Consider U_1, which is equal to
inf{s_n : n>1}. Now, this set consists of, 1, 1/2, 1/3 1/4, ... so the inf is 1/4 ????????????? but it goes ON ! 1/5, 1/6, 1/7 etc etc, the set {s_n : n > N} is NOT bounded. so how in the world are we suppose to generate U_N and V_N ?? We can't take infinity because that is suppose to be the limit inf/sup ! but then how in the world can we compare two U_N's. U_1 is suppose to be less then U_2, since U_N is a nondecreasing monotone sequence. But from my example, U_1 does not even exist because our set keeps decreasing.All my problems would be solved if the def was { s_n : N1 < n < N2 !!!!!!!!!!!!!!!}

6. In the example above, the sequence for V_N works out fine, because the set is bounded from below.

7. I think you need to relearn when sup and inf mean. The fact that $s_n$ if infinite doesn't matter.
In your example $U_n=0$ for all n, so you do have a nondecreasing sequence. Also you would have that $V_n=\frac{1}{n}$.
Try thinking of it this way: Removing terms can only decrease the upper bound and increase the lower bound. Do you see why? If not try working with simple finite sets fist before moving on to more complicated ones.

8. Originally Posted by putnam120
I think you need to relearn when sup and inf mean. The fact that $s_n$ if infinite doesn't matter.
In your example $U_n=0$ for all n, so you do have a nondecreasing sequence. Also you would have that $V_n=\frac{1}{n}$.
Try thinking of it this way: Removing terms can only decrease the upper bound and increase the lower bound. Do you see why? If not try working with simple finite sets fist before moving on to more complicated ones.
Be careful now. "lim sup" of a sequence is NOT the "sup" of the set forming the sequence. The "lim sup" of a sequence is the sup of the set of subsequential limits. (I was surprised at Zero266's original statement that he was to prove that "the limit sup s_n is always less then or equal to the limit s_n". If a sequence HAS a limit then all subsequences converge to that limit so in that case, lim sup s_n= lim inf s_n= lim s_n.)

For a decreasing sequence like {1/n}= {1, 1/2, 1/3, 1/4, ...}, the sup is, of course, 1 but since it has a limit, 0, the lim sup 1/n is 0.

9. Yes i know that there is a difference between limsup and sup (similarly for inf). I guess it it worth asking; did you [Zero266] mean to define

V_N := SUP {s_n: n>N}
or
V_N := LIMSUP {s_n: n>N}?

If you meant the latter then V_n is not a decreasing sequence in the way you are looking for. Instead it is constant.

10. Firstly, thanks for all the help. I'm starting to see the light. I define V_N = SUP {s_n: n> N}. After doing some examples, i discovered an interesting phenomena. When s_n = 1/n, U_N was the constant value 0, which as you mentioned, is indeed non-decreasing. In that same sequence, V_N was a non increasing sequence since each term got smaller and smaller.

If given a sequence s_n, does this mean that either U_N or V_N must be constant for all N? In other words, one of them is a constant and the other is an actual monotone sequence? Can you give me a specific seq where both U_N and V_N are montone sequences and where neither is a constant?

the point im trying to get at is understanding the DEF of sup/inf {s_n : n> N}. The part of the definition that confuses me is taking n > N. It seems that n is bounded from one side only, so either U_N or V_N must be constant for all N. A lot of my confusing would disappear if we took the interval say, N1 < n < N2 as oppose to n > N.

11. I'm not sure I really understand the root of the confusion, but I'll try to address some of these last points you've brought up.

If given a sequence s_n, does this mean that either U_N or V_N must be constant for all N? In other words, one of them is a constant and the other is an actual monotone sequence? Can you give me a specific seq where both U_N and V_N are montone sequences and where neither is a constant?
So to get it straight, $V_N=\displaystyle\sup_{n>N}{s_n}$ and $U_N=\displaystyle\inf_{n>N}{s_n}$ correct?

Consider $s_n=1 + (-1)^n\dfrac{1}{n}$. Then it should be easy so see $V_{2N-1}=V_{2N}>V_{2N+1}$ and $U_{2N-1}=U_{2N}>U_{2N+1}$ for all $N\geq 1$. So both are monotone.

the point im trying to get at is understanding the DEF of sup/inf {s_n : n> N}. The part of the definition that confuses me is taking n > N. It seems that n is bounded from one side only, so either U_N or V_N must be constant for all N. A lot of my confusing would disappear if we took the interval say, N1 < n < N2 as oppose to n > N
This is a somewhat subtle point. Ultimately, we want to say that the infimum of $V_n$ (respectively supremum of $U_n$) is the limsup (respectively liminf) of the sequence $\{s_n\}$, so it makes sense that we require infinitely many terms.

So why does this definition use the condition $n>N$?
Well, how do we assess the behavior of sequences (think convergence)? We throw out some finite number of terms and then try to show some sort of nice property (such as closeness to a proposed limit), right? So this is following the same paradigm.

So now to address the "Why do I care" feeling I am getting...

What does $\alpha=\displaystyle \inf_N V_N$ mean? By the definition of infimum, for every $\epsilon>0$ there is an $N(\epsilon)$ such that $V_{N(\epsilon)}\in[\alpha,\alpha+\epsilon)$. But that means that for all $n>N(\epsilon)$, $s_n<\displaystyle \sup_{n>N(\epsilon)}s_n<\alpha+\epsilon$. Thus there are only a finite number of terms exceeding $\alpha+\epsilon$ for any $\epsilon>0$! (that is, if $n\leq N(\epsilon)$, it is possible for $s_n>\alpha+\epsilon$) Put in the paradigm, the statement is "for all epsilon greater than zero, there exists N such that for all n>N $s_n$ is bounded above by $\alpha+\epsilon$".

This is what allows us to ultimately show that $\lim\sup s_n=\alpha=\displaystyle \inf_N \sup_{n>N} s_n$. This may seem like juggling symbols, but we can make great use of the bold statement to get a more tangible description of the limsup. That is, if $\alpha$ is a subsequential limit of $\{s_n\}$, then $\alpha=\lim\sup s_n$ if and only if for all $\epsilon>0$, there are only finitely many terms exceeding $\alpha+\epsilon$.