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series converges or diverges?

 
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Jan13-08, 08:29 PM   #1
 

series converges or diverges?


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.

am i on the right track, please help.
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Jan13-08, 11:28 PM   #2

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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?
Jan14-08, 02:51 AM   #3
 
Hint: try applying your logic to n/n.
Jan14-08, 05:27 AM   #4
 
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series converges or diverges?


Quote by rcmango View Post
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.
No, you don't know that! You only know that the limit is 0.

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

am i on the right track, please help.
No!
Jan14-08, 11:32 PM   #5
 
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.
Jan14-08, 11:44 PM   #6

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Quote by rcmango View Post
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.
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.
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