Determining Convergence of ((sin(n))^4)/(1+n^2)

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Discussion Overview

The discussion focuses on determining the convergence of the series \(\frac{(\sin(n))^4}{1+n^2}\) using the limit comparison test. Participants explore the behavior of the function as \(n\) approaches infinity and its implications for convergence.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests using the limit comparison test by dividing \(\frac{(\sin(n))^4}{1+n^2}\) by \(\frac{1}{1+n^2}\), leading to the consideration of \((\sin(n))^4\) as \(n\) approaches infinity.
  • Another participant confirms that \((\sin(n))^4\) is always between 0 and 1 and presents an inequality showing that \(\frac{(\sin(n))^4}{1+n^2} < \frac{1}{n^2}\) for all \(n > 0\), noting that \(\frac{1}{n^2}\) converges.
  • A later reply reiterates the earlier points and suggests that the range of \((\sin(n))^4\) being between 0 and 1 supports the conclusion of convergence, while also correcting the phrasing of a previous statement.

Areas of Agreement / Disagreement

Participants express similar ideas regarding the convergence of the series, but there is no explicit consensus on the reasoning behind the convergence, as some aspects remain debated.

Contextual Notes

The discussion does not resolve the nuances of the limit comparison test application or the implications of the behavior of \((\sin(n))^4\) as \(n\) approaches infinity.

hivesaeed4
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Suppose I want to determine the convergence of ((sin(n))^4)/(1+n^2) using limit comparison test. I divide it by 1/(1+n^2). All that remains is (sin(n))^4. Now as the limit goes to infinty, the range of values (sin(n))^4 can give is 0 to 1. Now it gives many more values above zero then at zero, so is that why we declare it a convergent series?
 
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As you point out, (sin(n))^4 is always between 0 and 1.

therefore

0 <= ((sin(n))^4)/(1+n^2) < 1/(1+n^2) < 1/n^2 for all n>0.

1/n^2 converges. So we're home and dry.
 
hivesaeed4 said:
Suppose I want to determine the convergence of ((sin(n))^4)/(1+n^2) using limit comparison test. I divide it by 1/(1+n^2). All that remains is (sin(n))^4. Now as the limit goes to infinty, the range of values (sin(n))^4 can give is 0 to 1.


*** Perhaps you meant "...can give is BETWEEN 0 and 1"...? ****


Now it gives many more values above zero then at zero, so is that why we declare it a convergent series?


Well, you almost have it : \frac{\sin^4(n)}{1+n^2}\leq\frac{1}{n^2} , which is clearly convergent.

DonAntonio
 
Lots of Thanks.
 

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