Show convergence using comparison test on sin(1/n)

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Homework Help Overview

The discussion revolves around testing the convergence of the series sin(1/n) using the comparison test. Participants are exploring the properties of the function and its relationship to known series.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to establish an inequality for the comparison test, with initial attempts suggesting sin(1/n) > 1/n, which is questioned by others. There is a focus on finding a correct comparison function and clarifying the nature of the inequalities involved.

Discussion Status

The discussion is ongoing, with participants providing insights and questioning assumptions about the inequalities. Some guidance has been offered regarding the approximation of sin(1/n) for large n, suggesting a potential path forward for establishing a valid comparison.

Contextual Notes

Participants are navigating the constraints of the comparison test and the accuracy of their inequalities. There is an acknowledgment of the need for correct assumptions to determine convergence or divergence.

ciarax
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Homework Statement




a) Test the following series for convergence using the comparison test
:
sin(1/n)

Explain your conclusion.

Homework Equations





The Attempt at a Solution


i must show f(x)<g(x) in order for it to converge other wise divergence.

g(x) = 1/n

sin(1/n) > 1/n always therefore the series is divergent.
is this correct?
 
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ciarax said:
sin(1/n) > 1/n

How did you obtain this inequality?? It isn't correct.
 
it was an attempt. i was looking back on my notes and thought that might be right :/ do you have any idea how to complete this?
 
The inequality sin(1/n) > 1/n might not hold, but it can be modified in something that does hold.
 
does this prove that it is divergent? or how do i go about proving if its convergent/divergent using the comparison test?
 
For large n, sin(1/n) ≈ 1/n. So use that to find your inequality.
 
does that give sin(1/n) > 1/2n?
 
It does for me.
 
so using that its divergent? thanks :)
 

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