Choosing Between the Ordinary & Limit Comparison Test

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SUMMARY

The discussion focuses on the choice between the Ordinary Comparison Test and the Limit Comparison Test in series convergence analysis. The Limit Comparison Test is applied to the series SUM n_infinity (3n - 2)/(n^3 - 2n^2 + 11), where the largest degree of terms is identified as 3/n^2. The reasoning behind this choice is clarified by noting that the relevant term in the numerator is 3n, leading to the simplification of 3n/n^3 = 3/n^2. Additionally, it is established that 3/n^2 converges to the p-series with p = 2.

PREREQUISITES
  • Understanding of series convergence tests, specifically the Ordinary Comparison Test and Limit Comparison Test.
  • Familiarity with polynomial degrees in rational functions.
  • Knowledge of harmonic series and p-series.
  • Basic algebraic manipulation skills for simplifying expressions.
NEXT STEPS
  • Study the properties and applications of the Ordinary Comparison Test in series convergence.
  • Learn about the Limit Comparison Test in detail, including its conditions and examples.
  • Explore p-series and their convergence criteria, focusing on different values of p.
  • Practice problems involving the identification of dominant terms in rational functions.
USEFUL FOR

Students studying calculus, particularly those focusing on series convergence, as well as educators teaching these concepts in mathematics courses.

rcmango
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Homework Statement



First questions is: How to choose between using the ordinary comparison test, or using the limit comparison test?



Homework Equations





The Attempt at a Solution



then, for these two problems below, i decide to use the limit comparison test:

SUM n_infinity (3n - 2)/(n^3 - 2n^2 + 11) then its said to look for the largest degree of terms in the numerator and denominator and divide through which is to divide by: 3/n^2

but why is 3/n^2 the largest term, when there is a n^3 in there, why not 3/n^3 ?

=======================

also, i see that 1/n belongs to the harmonic series, but what series does 3/n^2 belong to?

thankyou.
 
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Because the relevant term in the numerator is 3*n. 3*n/n^3=3/n^2.
 

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