Limit Comparison Test for series

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SUMMARY

The discussion focuses on the application of the Limit Comparison Test to the series sum 1/sqrt(3n-2) from n=1 to infinity. The comparison series used is sum 1/sqrt(n) from n=1 to infinity, which is known to be a divergent p-series. The user initially miscalculated the limit, arriving at 0, but later corrected this to find that the limit is actually 1/sqrt(3), confirming that the original series also diverges.

PREREQUISITES
  • Understanding of the Limit Comparison Test in calculus
  • Familiarity with p-series and their convergence properties
  • Knowledge of limits and their evaluation
  • Basic concepts of infinite series
NEXT STEPS
  • Study the Limit Comparison Test in detail with examples
  • Explore the properties of p-series and their convergence criteria
  • Practice evaluating limits involving square roots and rational functions
  • Investigate other convergence tests for infinite series, such as the Ratio Test and Root Test
USEFUL FOR

Students and educators in calculus, mathematicians analyzing series convergence, and anyone looking to deepen their understanding of infinite series and convergence tests.

bcjochim07
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I have been given the series
sum 1/sqrt(3n-2) from n=1 to infinity

I am supposed to use the limit comparison test, and the comparison series my book suggests is sum 1/sqrt n from n=1 to infinity, which I know is a divergent p series

However, when I take the limit of one divided by the other I come up with 0.

lim n -> infinity for (sqrt n)/(sqrt 3n-2) = 0

What am I doing wrong?
 
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Nevermind, I figured it out.

This Limit Comparison test shows that the series diverges because the limit is 1/sqrt3.
 

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