# Series converges or diverges?

1. Jan 13, 2008

### rcmango

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

n = 1 E infinity (n * sin (1/n))

2. Relevant equations

geometric series test?

3. The attempt at a solution

It looks like a geometric series.

i know that sin 1/n = 0 by squeeze theorem.

n * 0 will always be 0.

Last edited: Jan 13, 2008
2. Jan 13, 2008

### Dick

rcmango, you really have to work on your notation. Is the 'E' supposed to be a sigma? If so then the nth term of your series tends to one. Can it converge?

3. Jan 14, 2008

### zhentil

Hint: try applying your logic to n/n.

4. Jan 14, 2008

### HallsofIvy

Staff Emeritus
No, you don't know that! You only know that the limit is 0.

But none of your terms is n* 0 so that is irrelevant.

No!

5. Jan 14, 2008

### rcmango

Okay, thanks Dick, ya limit is 1.

i could use the comparison test, compare to 1/n
divide an / bn and get 1 by the nth term series. Thus diverging because it doesn't = 0.

thats what i made out of it. thanks alot.

6. Jan 14, 2008

### Dick

Pretty good. Except a ratio test giving you a limit of one doesn't tell you much. You could do a comparison with 1/n, that works. But it's still overkill. If the nth term of a series doesn't approach zero then it doesn't converge. Period. Ever.