Convergence of Series with Alternating Terms

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SUMMARY

The discussion centers on the convergence of the series E from 1 to infinity of [2+(-1)^n]/[n(n^1/2)]. Participants emphasize the importance of evaluating the series as a whole rather than splitting it into its numerator and denominator. The divergence test applied to the numerator indicates divergence, while the p-series test applied to the denominator confirms convergence. Ultimately, the conclusion is that one cannot assume the product of a diverging function and a converging function will diverge without further analysis.

PREREQUISITES
  • Understanding of series convergence tests, including divergence tests and p-series tests.
  • Familiarity with alternating series and their properties.
  • Knowledge of bounding techniques for series comparison.
  • Basic calculus concepts, particularly limits and infinite series.
NEXT STEPS
  • Study the properties of alternating series and the Alternating Series Test.
  • Learn about the Comparison Test for series convergence.
  • Explore bounding techniques for series to simplify convergence analysis.
  • Investigate the implications of combining converging and diverging series.
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Students studying calculus, particularly those focused on series convergence, as well as educators looking for effective teaching strategies for series analysis.

mvpshaq32
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Homework Statement


E from 1 to infinity [2+(-1)^n]/[n(n^1/2)]


Homework Equations



We only have had learned comparison tests, power series, and divergence tests.

The Attempt at a Solution



I decided to split the function into its numerator multiplied by its denominator:
2+(-1)^n * 1/[n(n^1/2)]
Then I perform the divergence test for the numerator in which the limit doesn't exist, meaning it diverges and the p-series test for the denominator whose ratio is 3/2 which means it converges. I then made the assumption that a diverging function multiplied by a converging function would diverge. But I don't know if I am allowed to do this.
 
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notice that
1 <= [2+(-1)^n] <= 3
 
mvpshaq32 said:
I decided to split the function into its numerator multiplied by its denominator...

I then made the assumption that a diverging function multiplied by a converging function would diverge. But I don't know if I am allowed to do this.
No, you can't do this. You need to look at the term as a whole.

You probably find the numerator confusing. A good tactic is to replace something that looks complicated with a simple quantity that either bounds it from above or from below, depending on what you want to prove. Then show the simplified series converges or diverges. Use lanedance's hint.
 

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