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Math Help - Convergence tests

  1. #1
    Junior Member
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    Oct 2010
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    Convergence tests

    When I'm testing \sum_{n = 0}^\infty \cos \frac{1}{n} for convergence, how come I can apply the nth term test and show that it is divergent, i.e.

    \lim_{n \to \infty} \cos \frac{1}{n} = \cos 0 = 1,

    but when I'm testing \sin \frac{1}{n} , I can't apply the same reasoning, i.e.

    \lim_{n \to \infty} \sin \frac{1}{n} = \sin 0 = 0 \Rightarrow \sin \frac{1}{n} converges,

    but instead have to use the limit comparison test and using L'Hopital's Rule, ending up with an implication that \sin \frac{1}{n} is divergent
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  2. #2
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    Do you know the difference between necessary and sufficient conditions?
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  3. #3
    Member
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    You can only show divergence that way. \lim_{n\to \infty}a_n=0 is necessary for convergence, but convergence does not follow from it.
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