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Math Help - is the series Sigma(1 to oo) sin/n convergent or divergent?

  1. #1
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    is the series Sigma(1 to oo) sin/n convergent or divergent?

    omg
    Really OMG !!
    It seems easy but really i cant test it !!
    Am loosing my head :/

    Is this one really hard or am forgetting my informations about testing the series for convergence ??!



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  2. #2
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    Quote Originally Posted by TWiX View Post
    omg
    Really OMG !!
    It seems easy but really i cant test it !!
    Am loosing my head :/

    Is this one really hard or am forgetting my informations about testing the series for convergence ??!




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  3. #3
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    Thank you.
    Thanks for my god .. Its hard, I thought am loosing the informations
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  4. #4
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    As an additional info: we know that the partial sum for \sin n is bounded so \sum a_n\sin n converges by Dirichlet test as long as a_n is a decreasing sequence which tends to zero.
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  5. #5
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    Actually, I dont know this test, but i think its the test with 2^n or something like that, Right?
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  6. #6
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    No, that's related with the Condensation test.

    Look at wikipedia, and seek for Dirichlet test and Cauchy's condensation test.
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  7. #7
    No one in Particular VonNemo19's Avatar
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    Isn't it the case that -1\leq\sin(n)\leq1 ?

    And doesn't this imply that

    \sum_{n=1}^{\infty}\frac{\sin{n}}{n} can be directly compared to \sum_{n=1}^{\infty}\frac{(-1)^n}{n} which converges by the alternating series test?
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  8. #8
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    Quote Originally Posted by VonNemo19 View Post
    Isn't it the case that -1\leq\sin(n)\leq1 ?

    And doesn't this imply that

    \sum_{n=1}^{\infty}\frac{\sin{n}}{n} can be directly compared to \sum_{n=1}^{\infty}\frac{(-1)^n}{n} which converges by the alternating series test?
    OK \frac{sin(n)}{n}\geq\frac{-1}{n}
    But what about \frac{(-1)^n}{n} ?
    ohhh since (-1)^n = \pm 1
    Am not sure about inquality, sometimes this bigger and sometimes its smaller.
    You are telling me \frac{sin(n)}{n}\geq\frac{(-1)^n}{n} ?
    If the smaller converges, You cant decide that convergence of the bigger.
    This is what I know.
    Or you are telling me \frac{sin(n)}{n}\leq\frac{(-1)^n}{n} ?
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  9. #9
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    No, that's wrong.

    In order to be the Comparison test applied, the general term must be nonnegative.
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  10. #10
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    Quote Originally Posted by Krizalid View Post
    No, that's wrong.

    In order to be the Comparison test applied, the general term must be nonnegative.
    WOW.
    I said that there is something wrong somewhere

    But It is correct that \frac{sin(n)}{n}\leq\frac{(-1)^n}{n} ??
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  11. #11
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Krizalid View Post
    No, that's wrong.

    In order to be the Comparison test applied, the general term must be nonnegative.
    Alternating series...
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  12. #12
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    No, re-read what I said.

    You're using comparison test, and a_n=\frac{\sin n}{n} is not always positive, so you can't use a direct comparison test.
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  13. #13
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    Quote Originally Posted by Krizalid View Post
    No, re-read what I said.

    You're using comparison test, and a_n=\frac{\sin n}{n} is not always positive, so you can't use a direct comparison test.
    Sir I know that I cant use the comparison tests.
    am asking about the inquality itself !

    Is it right ?

    Lol you are commented on his reply.
    Sorry.
    But is the inquality right?
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  14. #14
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    No.
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    Quote Originally Posted by TWiX View Post
    WOW.
    I said that there is something wrong somewhere

    But It is correct that \frac{sin(n)}{n}\leq\frac{(-1)^n}{n} ??
    \frac{\sin{3}}{3} = 0.04704000

    \frac{(-1)^3}{3} = -0.333333 < 0.0470400

    For example.
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