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Thread: convergent/divergent series

  1. #1
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    convergent/divergent series

    test for convergence and/or divergence

    $\displaystyle \sum_{n =1}^{\infty}n sin (1/n)$ is convergent?

    i'm not sure what to do when it comes to trig functions in series.. integral test? comparison test?
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  2. #2
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    Use the divergence test: If $\displaystyle \lim_{n \to \infty} a_n \neq 0$, then $\displaystyle \sum a_n$ diverges

    Here, consider: $\displaystyle \lim_{n \to \infty} n \sin \tfrac{1}{n} = \lim_{n \to \infty} \frac{\sin \tfrac{1}{n}}{\tfrac{1}{n}}$

    Let $\displaystyle \theta = \tfrac{1}{n}$. As $\displaystyle n \to \infty$, $\displaystyle \theta \to 0$.

    So our limit becomes the familiar: $\displaystyle \lim_{\theta \to 0} \frac{\sin \theta}{\theta}$
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  3. #3
    MHF Contributor matheagle's Avatar
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    Always first check whether the terms go to zero. If not, the series diverges. That's the situation here. There's no need for L'Hopital's Rule either. Let m=1/n...

    $\displaystyle n\sin (1/n)={\sin m\over m}\to 1 $ as $\displaystyle m\to 0$.
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