Limit superior & limit inferior of a sequence

In summary: HTHPetekI'm sorry, but I don't understand your notation at all.V_{N'} = {a_n: n \geq N'}What does this mean? Does it mean the set of all an such that n≥N'?What is v_{N'}? Is it the same as lim sup{an: n≥N'}?And what does it mean to say that V_1 \supseteq V_2 \supseteq V_3 \supseteq ..., andv_1 \geq v_2 \geq v_3 \geq ...Could you explain in words? I am really really confused.Any help is much appreciated
  • #1
kingwinner
1,270
0

Homework Statement


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

Theorem 1:
If lim an = a exists, then lim sup an = lim inf an = a.
n->∞

Theorem 2:
If lim sup an = lim inf an = a, then
lim an exists and equals a.
n->∞

Homework Equations


N/A

The Attempt at a Solution


I was trying to see why theorems 1 & 2 are true.
How can we prove these rigorously?

I wrote down all the definitions, but still I don't know how to prove theorems 1 and 2.

Let an be a sequence of real numbers. Then by definition, an->a iff
for all ε>0, there exists an integer N such that n≥N => |an - a|< ε.

Also, lim sup an is defined as
lim sup{an: n≥N}
N->∞
(similarly for lim inf)

Any help is much appreciated! :)
 
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  • #2
kingwinner said:

Homework Statement


Fact:
Let a=lim sup an.
Then for all ε>0, there exists N such that if n≥N, then an<a+ε
For Theorem 2 you also need:
Let a= lim inf an.
Then for all ε> 0 there exists N such that if n≥N, then an> a-ε
Theorem 1:
If lim an = a exists, then lim sup an = lim inf an = a.
n->∞
If the sequence converges to a, the every subsequence converges to a so the set of all "subsequential limits" is simply {a}.

Theorem 2:
If lim sup an = lim inf an = a, then
lim an exists and equals a.
n->∞
Since a= lim sup an and lim inf an, you have both of the above so that, for any ε> 0 there exist N such that if n≥ N then a-ε< an< a+ε.

Homework Equations


N/A

The Attempt at a Solution


I was trying to see why theorems 1 & 2 are true.
How can we prove these rigorously?

I wrote down all the definitions, but still I don't know how to prove theorems 1 and 2.

Let an be a sequence of real numbers. Then by definition, an->a iff
for all ε>0, there exists an integer N such that n≥N => |an - a|< ε.

Also, lim sup an is defined as
lim sup{an: n≥N}
N->∞
(similarly for lim inf)

Any help is much appreciated! :)
 
  • #3
1) About theorem 1, I haven't seen the version of the definition that you gave for lim sup and lim inf. My version of the definition is sipmly:
lim sup an is defined as
lim [sup{an: n≥N}]
N->∞

lim inf an is defined as
lim [inf{an: n≥N}]
N->∞

How can we prove theorem 1 directly by using these definitions (and the definition of "limit")?



2) Suppose a=lim sup an= lim inf an.
Then for all ε>0, there exists N1 such that if n≥N1, then an<a+ε
and for all ε> 0 there exists N2 such that if n≥N2, then an> a-ε
Should I take N=max{N1,N2}? so that n≥N => an<a+ε and an> a-ε, i.e. |an -a|< ε, and therefore an->a.
I think in general we should assume that N1 and N2 may be different (i.e. not necessarily the same N). Is this the correct way to prove theorem 2?


Can somebody help me, please?
Any help is much appreciated! :)
 
Last edited:
  • #4
Here are some hints for proving Theorem 1:

1. I'll assume that you're working over the real numbers, not the extended reals where values of + or - infinity are allowed.

2. Show (or simply observe, since it's easy) that lim inf [itex]a_n \leq[/itex] lim sup [itex]a_n[/itex].

3. Show that |[itex]a_n[/itex] - a| < [itex]\epsilon[/itex] implies that [itex]a_n[/itex] < a + [itex]\epsilon[/itex].

4. What, then, can you say about lim sup [itex]a_n[/itex] in terms of a and [itex]\epsilon[/itex]?

5. What does the relation in #4 imply about lim sup [itex]a_n[/itex] and a?

6. Using similar arguments, derive an analogous relation between lim inf [itex]a_n[/itex] and a.

7. Use the observation in #2 above to finish the proof.

HTH

Petek
 
  • #5
Thanks for your hints, but still I don't know how to prove theorem 1 (I have no idea how to prove 2,4,5).

Here is a proof of theorem 1 from my notes:
an->a
=> for all ε>0, there exists N such that if n≥N => |an-a|<ε
So n≥N => a-ε < an < a+ε
=> for all N' ≥N, a-ε ≤ sup{an: n ≥ N' } ≤ a+ ε
Thus, sup{an: n≥N} -> a as N->∞

==============================

But I really have absolutely no idea why the last two lines (highlighted in blue) are true. What is the point of introducing N' ? And why is it true that for all N' ≥N, a-ε ≤ sup{an: n ≥ N' } ≤ a+ ε?

Could someone please explain?
 
  • #6
for all n≥N, a-ε < an < a+ε
=> for all N' ≥N, a-ε ≤ sup{an: n ≥ N' } ≤ a+ ε

Why is this implication true? (particularly the lower bound)

Help...I am totally confused. Could someone kindly explain?
Any help is much appreciated! [I'm dying on this proof :( ]
 
  • #7
kingwinner said:
for all n≥N, a-ε < an < a+ε
=> for all N' ≥N, a-ε ≤ sup{an: n ≥ N' } ≤ a+ ε

Why is this implication true? (particularly the lower bound)

Help...I am totally confused. Could someone kindly explain?
Any help is much appreciated! [I'm dying on this proof :( ]

I'll try to explain this. To simplify the notation, define

[itex]V_{N'}[/itex] = {[itex]a_n[/itex]: n [itex]\geq[/itex] N'}

and

[itex]v_{N'}[/itex] = sup {[itex]V_{N'}[/itex]}

Then [itex]V_1 \supseteq V_2 \supseteq V_3 \supseteq ...[/itex], and
[itex]v_1 \geq v_2 \geq v_3 \geq ...[/itex]

(The second string of inequalities follows from the rule that if S and T are subsets of the reals and S [itex]\supseteq[/itex]T, then sup S [itex]\geq[/itex] sup T, which you should have already covered in your class.)

Now, using the above notation, we have to show that

[itex]a - \epsilon \leq v_{N'} \leq a + \epsilon[/itex]

For the right inequality, we have that [itex]a_n < a + \epsilon[/itex], so [itex]a + \epsilon[/itex] is an upper bound for [itex]V_{N'}[/itex]. Since [itex]v_{N'}[/itex] is the least upper bound for these sets, it follows that [tex]v_{N'} \leq a + \epsilon[/itex], as required.

To get the left inequality, argue as above, but use inf instead of sup. Define [itex]u_{N'}[/itex] = inf {[itex]a_{N'}: N' \geq N[/itex]}. Show that [itex]a - \epsilon \leq u_{N'}[/itex] similar to the above. Finally, observe that inf [itex]u_{N'} \leq [/itex]sup [itex]v_{N'}[/itex] and you get the left inequality.

HTH

Petek
 

FAQ: Limit superior & limit inferior of a sequence

What is the definition of limit superior and limit inferior of a sequence?

The limit superior of a sequence is the largest value that the sequence approaches as the index approaches infinity. The limit inferior is the smallest value that the sequence approaches as the index approaches infinity.

How are limit superior and limit inferior related to the convergence of a sequence?

If a sequence has a limit superior and limit inferior that are equal, then the sequence is said to converge. If the limit superior is greater than the limit inferior, the sequence is said to diverge.

Can the limit superior and limit inferior of a sequence be infinite?

Yes, the limit superior and limit inferior can both be infinite. This occurs when the sequence has unbounded values and does not converge.

How do you calculate the limit superior and limit inferior of a sequence?

To calculate the limit superior, you take the supremum (least upper bound) of the set of all limits of subsequences of the original sequence. To calculate the limit inferior, you take the infimum (greatest lower bound) of the set of all limits of subsequences.

Can a sequence have a limit superior or limit inferior if it does not converge?

Yes, a sequence can still have a limit superior and limit inferior even if it does not converge. The values of the limit superior and limit inferior will depend on the behavior of the sequence as the index approaches infinity.

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