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Math Help - [SOLVED] The difference of the harmonic series and 1/n comparsion

  1. #1
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    [SOLVED] The difference of the harmonic series and 1/n comparsion

    I recently had this classic example to determine if it converged or diverged
    [MATH\frac{sin(n)}{n}][/tex]

    I learned that you can use a comparison test using the squeeze theorem.

    \frac{-1}{n}\le\frac{sin(n)}{n}\le\frac{1}{n}

    Then you would take the limit of

    \lim_{n\to\infty}\frac{1}{n} which was go to 0.

    Because the function was greater than the original you can say the original converged as well to a limit of 0.

    However, how would the harmonic series, diverge if you are basically determining the limit of the same thing

    Edit: Poor question, trying to mark as solved, please disregard, if misused the concept of series and sequence
    Last edited by RockHard; November 18th 2009 at 11:25 PM.
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  2. #2
    MHF Contributor chisigma's Avatar
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    In...

    http://www.mathhelpforum.com/math-he...eries-4-a.html

    ... it has been demonstrated that the series...

    \sum_{n=1}^{\infty} \frac{\sin n}{n} , \sum_{n=1}^{\infty} \frac{\cos n}{n} (1)

    ... are both convergent and also their sum has been computed. I don't know if it exists an easier way to arrive at the same results... probably it exists ...

    Kind regards

    \chi \sigma
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  3. #3
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    Thank you very much, it seems to me I am confusing the terms series and sequence perhaps, if there is a real difference
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