Does this series converge.

1. Jan 16, 2007

rcmango

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

Does the series infinity n(sin(1/n)) converge?
E
n=1

2. Relevant equations

n

and

sin(1/n)

3. The attempt at a solution

The equation sin(1/n) looks familiar, maybe i could use the squeeze theorem?

something like -1 <= sin(1/n) <= 1

i'm not sure where to go after this though, or if i'm even on the right track?
thankyou.

2. Jan 16, 2007

arildno

First check:
Does the general term go to zero as n goes to infinity?

3. Jan 16, 2007

Gib Z

Yes You could use the squeeze theorem and get -|x|< sin(1/x) < |x|, but that doesn't seem the help you much. Do what arildno said, maybe a few other convergence tests, try using Taylor Series? I'd say it does, but can't show you right now.

4. Jan 16, 2007

dextercioby

Since this is not an alternating series, then only answering the question posed by Arildno is enough to solve the problem.

Daniel.

5. Jan 16, 2007

lkh1986

There is a general formula: as x approaches 0, (sin x)/x approaches 1. So you can apply this in your question. Let n=1/x, then you modify the equation, then you will get the answer that the series diverge, since the series does not have a sum.

6. Jan 16, 2007

rcmango

Last edited: Jan 16, 2007
7. Jan 16, 2007

rcmango

yes. i would say as n goes to infinity, it approaches 0.
the reasoning is because starting with small numbers it gets larger around .01745...

don't know where to go from here.

thanks for all the feedback, i found it all useful.

8. Jan 16, 2007

d_leet

Numbers at a certain point don't really tell you much about the behavior at infinity. Take the limit of the terms as n goes to infinity, if it isn't 0 what can you say about the series?

9. Jan 17, 2007

rcmango

that it diverges.

10. Jan 17, 2007

yes it does