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

In summary, the conversation is about testing the convergence of a given series using the comparison test. The discussion includes finding an inequality between the given series and a known series, and using this to determine if the given series is convergent or divergent. The final conclusion is that the given series is divergent based on the comparison with the known series.
  • #1
ciarax
6
0

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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  • #2
ciarax said:
sin(1/n) > 1/n

How did you obtain this inequality?? It isn't correct.
 
  • #3
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?
 
  • #4
The inequality sin(1/n) > 1/n might not hold, but it can be modified in something that does hold.
 
  • #5
does this prove that it is divergent? or how do i go about proving if its convergent/divergent using the comparison test?
 
  • #6
For large n, sin(1/n) ≈ 1/n. So use that to find your inequality.
 
  • #7
does that give sin(1/n) > 1/2n?
 
  • #8
It does for me.
 
  • #9
so using that its divergent? thanks :)
 

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

1. What is the comparison test for showing convergence of a series?

The comparison test is a method used to determine if a series converges or diverges by comparing it to a known convergent or divergent series.

2. How does the comparison test work?

The comparison test works by comparing the given series to a known series whose convergence or divergence is already known. If the given series is smaller than the known convergent series, then it also converges. If the given series is larger than the known divergent series, then it also diverges.

3. Why is the comparison test used for showing convergence of sin(1/n)?

The comparison test is used for showing convergence of sin(1/n) because it is a convenient way to compare the series to a known convergent series, such as 1/n, which has a known limit of 0 as n approaches infinity.

4. Can the comparison test be used to show divergence?

Yes, the comparison test can be used to show divergence if the given series is larger than a known divergent series, such as 1/n, which has a known limit of 0 as n approaches infinity.

5. Is the comparison test always accurate in determining convergence?

No, the comparison test is not always accurate in determining convergence. It can only be used to determine if a series converges or diverges, but it does not provide information about the actual value of the convergence or divergence.

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