Determining convergence of a series

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SUMMARY

The forum discussion focuses on determining the convergence of the series \(\Sigma\frac{n+5}{\sqrt[3]{n^7+n^2}}\) from \(n=1\) to infinity. The initial application of the Test for Divergence yielded a limit of 0, indicating convergence. However, the Ratio Test was inconclusive, prompting suggestions to explore rationalizing the denominator or employing a Comparison Test for a definitive conclusion.

PREREQUISITES
  • Understanding of series convergence tests, including the Test for Divergence and Ratio Test.
  • Familiarity with the Comparison Test for series.
  • Knowledge of rationalizing techniques in calculus.
  • Basic algebraic manipulation skills for simplifying expressions.
NEXT STEPS
  • Study the Comparison Test in detail to apply it effectively to series.
  • Learn about rationalizing denominators in the context of series convergence.
  • Explore advanced convergence tests such as the Limit Comparison Test.
  • Review examples of series that converge and diverge to solidify understanding.
USEFUL FOR

Students studying calculus, particularly those focused on series convergence, as well as educators seeking to enhance their teaching methods in this area.

grothem
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Homework Statement


Determine if the following is convergent or divergent
[tex]\Sigma\frac{n+5}{\sqrt[3]{n^7+n^2}}[/tex] n from 1 to infinity


Homework Equations


Test for divergence came up with limit = 0 so I know it converges.


The Attempt at a Solution


Ratio test came up inconclusive. Should I try rationalizing the denominator?

 
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use a comparison test, and then u can see if diverges or converges
 

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