Simple Convergence / Divergence Calc 2

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SUMMARY

The discussion centers on determining the convergence or divergence of the series defined by the summation from n=1 to infinity of 1 / sqrt(n^3 + 2n). The consensus is that the Comparison Test is the most effective method for this problem, as the series can be compared to the p-series 1/n^(3/2), which converges since p > 1. The integral test is deemed less practical due to the complexity of direct application without approximations.

PREREQUISITES
  • Understanding of series convergence tests, specifically the Comparison Test and Integral Test.
  • Familiarity with p-series and their convergence criteria.
  • Basic knowledge of limits and asymptotic behavior of functions.
  • Ability to manipulate and analyze mathematical expressions involving square roots and polynomials.
NEXT STEPS
  • Study the Comparison Test in detail, focusing on its application in series convergence.
  • Learn about p-series and the conditions under which they converge or diverge.
  • Explore the Integral Test and its limitations when applied to complex series.
  • Practice problems involving convergence tests to solidify understanding and application skills.
USEFUL FOR

Students studying calculus, particularly those tackling series convergence problems, as well as educators looking for examples of convergence tests in action.

Alex G
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Homework Statement



I have stared at this too long and do not know which test to approach it with, even writing it out. The problem is
State the Convergence or Divergence of the given series:

Summation n=1 to Infinity of 1 / sqrt (n^3 + 2n)

Homework Equations


I narrowed it down to possibly a comparison test, I can't really see an integral test coming in.
I know the answer in the back of the book says convergence
 
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Comparison test works well here since sqrt (n^3 + 2n) > n^3/2. The later being a p series with p >1.

Integral test would be hard to use directly without any approximations which can be done with the comparison test.
 
I thought the same, however, as I calculate out different and higher successive numbers for
1 / sqrt(n^3 + 2n), it's always smaller than 1 / n^(3/2) :(
 
Alex G said:
I thought the same, however, as I calculate out different and higher successive numbers for
1 / sqrt(n^3 + 2n), it's always smaller than 1 / n^(3/2) :(

If you want to prove convergence, isn't 1/sqrt(n^3+2n)<1/n^(3/2) the condition you want?
 
Alex G said:
I thought the same, however, as I calculate out different and higher successive numbers for
1 / sqrt(n^3 + 2n), it's always smaller than 1 / n^(3/2) :(

That's precisely what you want... ;)
 
/sigh I'm sorry, this is my 6th chapter review section ... I think I'm going to quit while ahead haha, thank you!
 

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