Simple Convergence / Divergence Calc 2

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Homework Help Overview

The discussion revolves around determining the convergence or divergence of the series defined by the summation from n=1 to infinity of 1 / sqrt(n^3 + 2n). The subject area is calculus, specifically focusing on series convergence tests.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the applicability of the comparison test and the integral test for the series in question. There is uncertainty regarding which test is most appropriate, with some participants expressing confusion about the behavior of the series as n increases.

Discussion Status

The conversation reflects a mix of attempts to apply the comparison test and questions about the conditions for convergence. Some participants acknowledge the answer provided in the textbook while grappling with their calculations and interpretations of the series' behavior.

Contextual Notes

Participants mention the challenge of the problem being part of a chapter review and express feelings of frustration with the material. There is a recognition of the difficulty in applying certain tests without approximations.

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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