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Thread: Limit Comparison Test of the series of sin(1/n)

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    Limit Comparison Test of the series of sin(1/n)

    What series do I compare $\displaystyle \sum_{n=1}^\infty \sin \frac 1n$ to when using the Limit Comparison test.
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    Re: Limit Comparison Test of the series of sin(1/n)

    Quote Originally Posted by MSUMathStdnt View Post
    What series do I compare $\displaystyle \sum_{n=1}^\infty \sin \frac 1n$ to when using the Limit Comparison test.
    $\displaystyle \sum_{n=1}^\infty \frac{ 1}{n}$
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    Re: Limit Comparison Test of the series of sin(1/n)

    But that won't be sufficient, because for large n, $\displaystyle \sin{\frac{1}{n}} < \frac{1}{n}$

    $\displaystyle \sum_{n=1}^\infty \frac{ 1}{n}$ diverges, but the series in the original post consists of smaller terms, so you can't use that as a direct comparison. You can however, use the fact that

    $\displaystyle \sin{\frac{1}{n}} > \frac{1}{2n}$

    and the fact that the below diverges:

    $\displaystyle \sum_{n=1}^\infty \frac{ 1}{2n}$

    In fact the 2 in the above series can be ANY number greater than 1.
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    Re: Limit Comparison Test of the series of sin(1/n)

    Quote Originally Posted by SworD View Post
    But that won't be sufficient, because for large n, $\displaystyle \sin{\frac{1}{n}} < \frac{1}{n}$

    @SworD, you did not read the question carefully.

    The question is about limit comparison not basic comparison.
    Some authors like Gillman call it ratio comparison.

    $\displaystyle \lim _{n \to \infty } \frac{{\sin \left( {\frac{1}{n}} \right)}}{{\frac{1}{n}}} = 1$
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    Re: Limit Comparison Test of the series of sin(1/n)

    Yes, the (ordinary) comparison test is if $\displaystyle 0 \le a_n \le b_n$ where $\displaystyle \sum_{n=1}^\infty{b_n}$ is a convergent series, then $\displaystyle \sum_{n=1}^\infty{a_n}$ converges. Or if $\displaystyle 0 \le b_n \le a_n$ where $\displaystyle \sum_{n=1}^\infty{b_n}$ is a divergent series, then $\displaystyle \sum_{n=1}^\infty{a_n}$ diverges.

    For the limit comparison test, if $\displaystyle \lim_{n \rightarrow \infty}\frac{a_n}{b_n}$ is finite and nonzero, then $\displaystyle \sum_{n=1}^\infty{a_n}$ converges if and only if $\displaystyle \sum_{n=1}^\infty{b_n}$ converges.

    So you need to compare $\displaystyle \sin(1/n)$ to 1/n, since $\displaystyle \lim_{n \rightarrow \infty}\frac{\sin(1/n)}{1/n}=1$ is finite and nonzero. Of course, 1/2n or 35/87n would also work - the limits would still be finite and nonzero.

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