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

1. Jan 9, 2012

ciarax

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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?

2. Jan 9, 2012

micromass

How did you obtain this inequality?? It isn't correct.

3. Jan 9, 2012

ciarax

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. Jan 9, 2012

micromass

The inequality sin(1/n) > 1/n might not hold, but it can be modified in something that does hold.

5. Jan 9, 2012

ciarax

does this prove that it is divergent? or how do i go about proving if its convergent/divergent using the comparison test?

6. Jan 9, 2012

Joffan

For large n, sin(1/n) ≈ 1/n. So use that to find your inequality.

7. Jan 9, 2012

ciarax

does that give sin(1/n) > 1/2n?

8. Jan 9, 2012

Joffan

It does for me.

9. Jan 9, 2012

ciarax

so using that its divergent? thanks :)