# Limit superior (lim sup) of a sequence

1. Jan 14, 2010

### kingwinner

Fact:
Let a=lim sup an.
Then for all ε>0, there exists N such that if n≥N, then an<a+ε
============================

1) What is the "intuitive meaning" of this fact? I cannot seem to make sense of it...

2) Let a=lim sup an.
Is it also true that:
For all ε>0, there exists N such that if n≥N, then

sup an<a+ε ?
n≥N

Why or why not?

Thanks for explaining!

Last edited: Jan 15, 2010
2. Jan 15, 2010

### HallsofIvy

The "lim sup" of a sequence is the least upper bound of the set of all subsequential limits of the sequence. IF, for any N, there were n> N such that an> a+ ε, then the subsequence of all such an would either not converge, contradicting the fact that lim sup is finite, or would converge to something larger than a, contradicting the fact that the lim sup is a.

No, of course, not! Another property of lim sup is that it is equal to $\lim_{n\ge N} a_n$, not less than that.

3. Jan 15, 2010

### breedencm

I don't understand what you mean HallsofIvy. Kingswinner's claim seems completely valid. Using his fact; for a = lim sup a_n, then for any e > 0 there exists and N such that for all n >= N, a_n < a + e.

So, considering that a_n < a + e for all n >= N. Then a + e is an upper bound, so clearly, sup a_n <= a + e, (the sup taking for all n >= N).

Kingswinner; Intiutively, HallsofIvy is absolutely correct, lim sup a_n, can be thought of the largest subsequential limit of any sequence a_n. However, it's usually best thought upon by using it's properties, and gaining experience. There are a few very nice features of this lim sup concept. (1) This limit _always_ exists (if we allow +/- infinity to be included into our definitions). So anytime you are considering a theorem for limits, instead of having to consider if a sequence converges or not, maybe you can just consider its lim sup. By doing this you are incorporating a larger class of sequences into your theorem, now it's not necessary to have a convergent sequence to apply your theorem (Reference the "ratio test" for series convergence as an example). Using lim sup would also give us somewhat of an 'upper bound' on the sequence, which is how it is used in the ratio test example. I hope this helps a little.

Last edited: Jan 15, 2010
4. Jan 15, 2010

### kingwinner

1) I'm sorry...I haven't learned the definition of lim sup that you gave...

My definition of limit superior is this:
lim sup an is defined as
lim sup{an: n≥N}
N->∞

How can we interpret the "Fact" using this version of the definition? Why does it make sense?

2) So is it true or not? Can sombody confirm?

Thank you!

Last edited: Jan 15, 2010
5. Jan 15, 2010

### breedencm

2) Let a=lim sup an.
Is it also true that:
For all ε>0, there exists N such that if n≥N, then

sup an<a+ε ?
n≥N.

Use this fact. This should be clear, by the definition of convergence. Also note that the existence of this limit is based on the fact that it is monotonically decreasing.

6. Jan 15, 2010

### HallsofIvy

You are right. I misread the formula. I thought he was saying a_n> a+ e!

7. Jan 16, 2010

### kingwinner

Statement:
Let a=lim sup an.
Then for all ε>0, there exists N such that if n≥N, then
sup an<a+ε ?
n≥N
======================

But does the statement above even make sense?

sup an = sup {aN,aN+1,aN+2,...} =bN
n≥N
This sequence of supremums are indexed by "N", i.e. the subscript of this sequence of sup's should be "N".

So the above statement is:
"For all ε>0, there exists N such that if n≥N, then
bN<a+ε ?"

It says that if n≥N, then "something" holds. But that "something" doesn't even depend on "n" (it does not even contain any "n" in it). Isn't this weird?? Does this mean that the statement above is wrong??

8. Jan 20, 2010

### a_Vatar

HallsofIvy,
I'm studying with Rudin’s Principles of Mathematical Analysis. They define lim sup exactly as sup of a set of all subsequential limits.
So let E be a set of all subsequential limits and let x = lim sup(an) = sup(E), then
if an > x = lim sup(an) for all but finitely many an, it is claimed that there's y in E such that y >= an > x contradicting definition of x...
Now this correlates with your explanation, but I still don't get it. The subsequence of an need not diverge but may "fluctuate" above x (e.g the given subsequence is bounded)... Don't see how there's necessarily another y in E for which an <= y.
Would I have to prove that we can extract a convergent subsequence from an? But what if an converges to x and still an > x for infinately many n???

9. Jun 23, 2010

### manoguru

Yes, the statement "For all ε>0, there exists N such that if n≥N, then bN<a+ε " is quite correct. To see this, we need to realize that bN is in fact a monotonically decreasing sequence (how?) and a = limN->infty bN is the lowest value that bN tends to approach. Since a+ε is greater than 'a', there is sure to be some values of bN such that a+ε ≥ bN ≥ a. Also since bN is monotonically decreasing, beyond some value of N, all subsequent values of bN will be less than a+ε.