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Math Help - [SOLVED] Series - Converge or diverge

  1. #1
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    [SOLVED] Series - Converge or diverge

    Hello,

    How can I show if these series are convergent or divergent?

    k = 1∞ cos((k^(3))/(3^(k))

    k = 1∞ ((-1)^(k))((sin(1/k))/(k))

    Appriciate any guidance and a detail explanation.

    Thank you
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  2. #2
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    recall |\cos x|\le1, then first series converges absolutely by direct comparison test by the geometric series.

    as for the second one, try Leibniz test.
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  3. #3
    Super Member Deadstar's Avatar
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    Quote Originally Posted by Krizalid View Post
    recall |\cos x|\le1, then first series converges absolutely by direct comparison test by the geometric series.

    as for the second one, try Leibniz test.
    Hmm...

    I thought the first one diverges...

    Doesn't \lim_{k \to \infty} \frac{k^3}{3^k} = 0 which would mean the sum just becomes 1+1+1+1+1+... (or near enough 1)...


    I just checked in Maple and you have...

    \sum_{k=1}^{n} \cos(\tfrac{3^k}{k^3}) tends to n-1 (or n-1.3 or something like that...) for n large
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  4. #4
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    The first one is pretty easy.

    You've got:

    <br />
\lim_{k \to \infty} cos(\frac{k^3}{3^k}) =cos( \lim_{k \to \infty} \frac{k^3}{3^k})=cos0=1 \neq 0

    and it is divergent. If it was 0 the test would've been inconclusive. (limit test)

    For the second one:

    you can see that \lim_{x->\infty}\frac{sin(\frac{1}{x})}{x}=0

    and it is decreasing (by drawing the graphic of the function \frac{sin(\frac{1}{x})}{x}.

    We can conclude that it is alternating series which converge.

    Regards.
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  5. #5
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    Quote Originally Posted by Krizalid View Post
    recall |\cos x|\le1, then first series converges absolutely by direct comparison test by the geometric series.

    as for the second one, try Leibniz test.
    I think the OP means \cos \left( \frac{k^3}{3^k} \right) (but it is very easy to misread it, even when reading it carefully).
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  6. #6
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    yes, i misread it, that's why my answer doesn't fit here.
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