Confirming an Inequality: Seeking Help

  • Context: Undergrad 
  • Thread starter Thread starter Square1
  • Start date Start date
  • Tags Tags
    Inequality
Click For Summary
SUMMARY

The discussion centers on the application of the comparison test in analyzing series convergence. The user seeks confirmation regarding the limit of b-subn equaling 0, which renders the comparison test inconclusive. Additionally, the user asserts that their simplified form of a-subn is valid and that their inequality holds true, despite the series diverging for large n, where the terms approximate 1/n.

PREREQUISITES
  • Understanding of series convergence tests, specifically the comparison test.
  • Familiarity with limits and their implications in series analysis.
  • Knowledge of divergent series, particularly those resembling harmonic series.
  • Basic mathematical notation and terminology related to sequences and series.
NEXT STEPS
  • Study the formal definition and application of the comparison test in series convergence.
  • Explore the implications of limits in series analysis, focusing on cases where limits equal 0.
  • Learn about divergent series and their characteristics, particularly the harmonic series.
  • Review mathematical proofs related to inequalities in series convergence.
USEFUL FOR

Mathematicians, students studying calculus or real analysis, and anyone interested in series convergence and divergence analysis.

Square1
Messages
143
Reaction score
1
Hey check out the attached picture. I think I solved the issue now but just to confirm perhaps...

If I were to continue here using the comparison test, is the only problem that b-subn limit equals 0, so the comparison test is inconclusive?

Otherwise, since our summation does not included n=0 (or other undefined values right?), my simplified form of a-subn holds, and my inequality is correct?

Thanks for any help.
 

Attachments

  • series pic.jpg
    series pic.jpg
    21.5 KB · Views: 407
Physics news on Phys.org
Your picture is hard to read. However, it seems you have a series where for large n, the terms are approximately 1/n, so the series diverges.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
30K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 25 ·
Replies
25
Views
4K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 49 ·
2
Replies
49
Views
6K
Replies
3
Views
3K