Proof of Divergence of a Series

Click For Summary
SUMMARY

The series \(\sum_{k=1}^{\infty}\sqrt[k]{k+1}-1\) diverges. The discussion highlights the challenge of proving that the \(n^{th}\) term is greater than \(\frac{1}{n}\). A key insight involves using the inequality \(n^n > (n+1)^{n-1}\) and applying logarithmic simplification. The series expansion of \(e^x\) is also referenced as a tool for analysis.

PREREQUISITES
  • Understanding of series convergence and divergence
  • Familiarity with logarithmic functions and properties
  • Knowledge of exponential functions and their series expansions
  • Basic skills in mathematical inequalities
NEXT STEPS
  • Study the properties of series convergence tests, particularly the Ratio Test
  • Explore the application of logarithmic inequalities in series analysis
  • Learn about the series expansion of \(e^x\) and its implications in calculus
  • Investigate advanced topics in asymptotic analysis and growth rates
USEFUL FOR

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

Arkuski
Messages
38
Reaction score
0
Prove that the series \displaystyle\sum_{k=1}^{\infty}\sqrt[k]{k+1}-1 diverges.

I thought that I could show the n^{th} term was greater than \frac{1}{n} but this is turning out to be more difficult than I imagined. Is there a neat proof that n^n>(n+1)^{n-1}?
 
Physics news on Phys.org
##(k+1)^\frac{1}{k}=\exp(\log((k+1)^\frac{1}{k}))##. Simplify the log a little and think about what the series expansion of ##e^x## looks like.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
29
Views
3K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K