Formal definition of a sequence limit which tends to infinity

Click For Summary
SUMMARY

The discussion centers on defining a sequence \(X_n\) that tends to infinity, specifically addressing the concept of limits in analysis. The correct definition states that for any large number \(M\), there exists an index \(N\) such that \(a_n > M\) for all \(n \geq N\). This contrasts with finite limits, where the epsilon-delta condition applies. The confusion arises from incorrectly attempting to apply the epsilon-delta definition to sequences with infinite limits.

PREREQUISITES
  • Understanding of sequences and limits in mathematical analysis
  • Familiarity with the epsilon-delta definition of limits
  • Basic knowledge of convergence criteria for sequences
  • Concept of infinity in mathematical contexts
NEXT STEPS
  • Study the formal definition of limits for sequences in mathematical analysis
  • Explore examples of sequences that converge to infinity
  • Learn about the differences between finite and infinite limits
  • Review convergence criteria and their applications in real analysis
USEFUL FOR

Students of mathematical analysis, educators teaching calculus or real analysis, and anyone seeking to deepen their understanding of sequences and limits in mathematics.

harmonie_Best
Messages
7
Reaction score
0

Homework Statement


hey there I have been given a question that asks me to define a sequence Xn which tends to infinity that has a limit that is infinity! I am so cofused. I would assume to use an adapted version of the epsilon delta condition?


Homework Equations



It's for Analysis!

The Attempt at a Solution



I attempted :
For all E > 0, there exists a delta >0 s.t. 0<!x!<delta then ?

Thanks!
 
Physics news on Phys.org
harmonie_ post: 2919712 said:

Homework Statement


hey there I have been given a question that asks me to define a sequence Xn which tends to infinity that has a limit that is infinity! I am so cofused. I would assume to use an adapted version of the epsilon delta condition?


Homework Equations



It's for Analysis!

The Attempt at a Solution



I attempted :
For all E > 0, there exists a delta >0 s.t. 0<!x!<delta then ?

Thanks!
There is no delta in the definition of convergence for a sequence that converges to a finite number. So for a sequence {an} that converges to L, the definition says that there is some number N such that |an - L| < epsilon, for all n >= N.

For a sequence whose limit is infinite, neither delta nor epsilon play a role in the definition. For this type of sequence, the definition is: For any large number M, there is a number N such that an > M for all n >= N.

The main idea here is that no matter how large a number M someone picks, there is an index N so that all the terms in the sequence past that index are larger than M.
 

Similar threads

  • · Replies 58 ·
2
Replies
58
Views
5K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
19
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K