Proving Divergence of Alternating Series with Limit Comparison Test

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SUMMARY

The discussion centers on proving the divergence of the alternating series \(\sum(-1)^{n-1} \frac{n-1}{n+1} \left(\frac{1}{\sqrt[n]{n}}\right)\). It is established that the limit \(\lim_{n \to \infty} \frac{n-1}{n+1} \left(\frac{1}{\sqrt[n]{n}}\right)\) is not zero, which is a necessary condition for divergence in alternating series. The participants confirm that while this condition suffices for regular series, additional considerations are required for alternating series. The conclusion emphasizes the importance of the limit of the terms approaching zero for convergence.

PREREQUISITES
  • Understanding of alternating series and their convergence criteria
  • Familiarity with the Limit Comparison Test in series analysis
  • Knowledge of limits and their properties in calculus
  • Basic proficiency in manipulating series and sequences
NEXT STEPS
  • Study the criteria for convergence and divergence of alternating series
  • Learn about the Limit Comparison Test and its applications in series
  • Explore examples of alternating series to solidify understanding
  • Investigate the implications of the terms' limits in series convergence
USEFUL FOR

Mathematics students, educators, and anyone studying series convergence, particularly those focusing on alternating series and limit comparison techniques.

Dell
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how do i prove that the following series diverges?

\sum(-1)n-1*\frac{n-1}{n+1}*(1/\sqrt[n]{n})

is it enough to say that

lim {n->inf} \frac{n-1}{n+1}*(1/\sqrt[n]{n}) is not 0 ?


i know for regular series this would be okay but what about series with alternating signs??

how else can i tbe proven??
 
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Yes, an alternating series converges if and only if the limit of the terms go to zero.
 
thanks, youve really helped me with this work,
 

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