Does \(\sum_{n=2}^{\infty}\frac{1}{(\ln n)^k}\) Converge for \(k > 1\)?

Click For Summary
SUMMARY

The series \(\sum_{n=2}^{\infty}\frac{1}{(\ln n)^k}\) converges for \(k > 1\). The divergence of \(\sum_{n=2}^{\infty}\frac{1}{\ln n}\) establishes a baseline for comparison. For \(k > 1\), the term \(\frac{n}{(\ln n)^k}\) approaches infinity, indicating that the series diverges. A formal proof can be constructed using L'Hôpital's rule and the comparison test with the harmonic series.

PREREQUISITES
  • Understanding of series convergence and divergence
  • Familiarity with L'Hôpital's rule
  • Knowledge of the harmonic series
  • Basic concepts of limits in calculus
NEXT STEPS
  • Study the comparison test for series convergence
  • Learn about L'Hôpital's rule in depth
  • Explore the properties of the harmonic series
  • Investigate series with logarithmic terms in greater detail
USEFUL FOR

Mathematicians, calculus students, and anyone studying series convergence, particularly those interested in logarithmic functions and their behavior in infinite series.

photis
Messages
7
Reaction score
0
It's easy to see that [tex]\sum_{n=2}^{\infty}\frac{1}{lnn}[/tex] does not converge. But what happens to [tex]\sum_{n=2}^{\infty}\frac{1}{(lnn)^k}[/tex] with k > 1 and why?

Can anybody help?
 
Physics news on Phys.org
i'm pretty sure it diverges for all k>1. this is because n/((lnn)^k) goes to infinity for all integer values of k. THink about this - use l'hospital's rule and definition of a limit for a formal proof using the comparison test with the harmonic series.
 
Thanks!
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 17 ·
Replies
17
Views
6K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K