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Math Help - [SOLVED] Infinite Series Convergence

  1. #1
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    [SOLVED] Infinite Series Convergence

    Hello, I have no idea how to show that the following infinite series diverge. I am supposed to compare then to the harmonic series... but I just don't get it.

    1. \sum ^\infty _{n=2} \frac{1}{(\log {n})^k}

    Hint \frac {1}{(\log {n})^k} > \frac {1}{n} for a sufficiently large n

    2. \sum ^\infty _{n=1} \sin {\frac {1}{n} }

    Hint \sin {\frac {1}{n}} > \frac {1}{2n} for a sufficiently large n

    Thanks in advance.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by akolman View Post
    Hello, I have no idea how to show that the following infinite series diverge. I am supposed to compare then to the harmonic series... but I just don't get it.

    1. \sum ^\infty _{n=2} \frac{1}{(\log {n})^k}

    Hint \frac {1}{(\log {n})^k} > \frac {1}{n} for a sufficiently large n

    2. \sum ^\infty _{n=1} \sin {\frac {1}{n} }

    Hint \sin {\frac {1}{n}} > \frac {1}{2n} for a sufficiently large n

    Thanks in advance.
    For the first one look here

    http://www.mathhelpforum.com/math-he...rgent-not.html

    For the second one consider \lim_{n\to\infty}\frac{\sin\left(\frac{1}{n}\right  )}{\frac{1}{n}}
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  3. #3
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    So is this correct?
    \lim_{n\to\infty}\frac{\sin\left(\frac{1}{n}\right  )}{\frac{1}{n}}<br />
=\lim_{u\to 0}\frac{\sin (u)}{u}=1 where u=\frac{1}{n}
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by akolman View Post
    So is this correct?
    \lim_{n\to\infty}\frac{\sin\left(\frac{1}{n}\right  )}{\frac{1}{n}}<br />
=\lim_{u\to 0}\frac{\sin (u)}{u}=1 where u=\frac{1}{n}
    Perfect....so we can assume that they share ____ by the ________ test
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