I cant understand this explanation of limsup

  • Thread starter Thread starter transgalactic
  • Start date Start date
  • Tags Tags
    Explanation
Click For Summary
SUMMARY

The discussion centers on the concept of limsup in the context of bounded sequences. It is established that for a bounded sequence \( x_n \), the limit superior is defined as \( \limsup x_n = \lim \left( \sup \{ x_k | k \geq n \} \right) \). The sequence \( \sup \{ x_k | k \geq n \} \) is non-increasing, leading to the conclusion that its limit is the infimum of the set \( \{ \sup \{ x_k | k \geq n \} | n \geq 0 \} \). An example using the sequence \( x_n = (-1)^n/n \) illustrates how the supremum changes as \( n \) increases, confirming the non-increasing nature of the sequence of suprema.

PREREQUISITES
  • Understanding of sequences and limits in real analysis
  • Familiarity with the concepts of supremum and infimum
  • Basic knowledge of bounded sequences
  • Ability to interpret mathematical notation and definitions
NEXT STEPS
  • Study the properties of bounded sequences in real analysis
  • Learn about the definitions and applications of supremum and infimum
  • Explore examples of limit superior and limit inferior in various sequences
  • Investigate the implications of non-increasing sequences in mathematical analysis
USEFUL FOR

Students and professionals in mathematics, particularly those studying real analysis, as well as educators seeking to clarify the concept of limit superior in bounded sequences.

transgalactic
Messages
1,386
Reaction score
0
regarding this definition

http://img515.imageshack.us/img515/5666/47016823jz1.gif

i was told that
Remember that if x_n is bounded then \limsup x_n = \lim \left( \sup \{ x_k | k\geq n\} \right).
The sequence, \sup \{ x_k | k\geq n\} is non-increasing, therefore its limits is its infimum.
Thus, \limsup x_n = \inf \{ \sup\{ x_k | k\geq n\} | n\geq 0 \}[/quote]


i can't understand the first part

why he is saying that
\sup \{ x_k | k\geq n\}
is not increasing.
you are taking a bounded sequence and you get one number
which is SUP (its least upper bound)
thats it.
no more members

??
 
Last edited by a moderator:
Physics news on Phys.org
But \left{x_k|k\ge n} is not a single sequence- it is a different sequence for every different n.

For example, if {x_n= (-1)^n/n}= {-1, 1/2, -1/3, 1/4, -1/5, ...} then<br /> sup{x_k|k\ge 1} is the largest of {-1, 1/2, -1/3, 1/4, -1/5, ...} which is 1/2. sup{x_k|k\ge 2} is the largest of {1/2, -1/3, 1/4, -1/5, ...}, again 1/2. sup{x_k|k\ge 3} is the largest of {-1/3, 1/4, -1/5, ...}, which is 1/4. Similarly, sup{x_k|k\ge 4} is also 1/4 but sup{x_k|k\ge 5} is 1/6, etc.
 

Similar threads

Replies
8
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
2K