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Series converges or diverges?

  1. Jan 13, 2008 #1
    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.
    Last edited: Jan 13, 2008
  2. jcsd
  3. Jan 13, 2008 #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?
  4. Jan 14, 2008 #3
    Hint: try applying your logic to n/n.
  5. Jan 14, 2008 #4


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    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.

  6. Jan 14, 2008 #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.
  7. Jan 14, 2008 #6


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    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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