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Math Help - How do I determine if the series converges?

  1. #1
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    How do I determine if the series converges?

    Its given that the series are as follows:


    ∑ (arctan k)/(1 + kČ)
    k=1
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  2. #2
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    Quote Originally Posted by KumaKuma View Post
    Its given that the series are as follows:


    ∑ (arctan k)/(1 + kČ)
    k=1
    0 < \frac{\pi}{4} < \frac{\arctan k}{1 + k^2} < \left( \frac{\pi}{2} \right) \, \frac{1}{k^2} < \frac{2}{k^2} for k > 0 ......
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    Quote Originally Posted by mr fantastic View Post
    0 < \frac{\pi}{4} < \frac{\arctan k}{1 + k^2} < \left( \frac{\pi}{2} \right) \, \frac{1}{k^2} < \frac{2}{k^2} for k > 0 ......
    Sorry...im poor in calculus..not very clear with your provided answer, can you explain further??
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  4. #4
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    Quote Originally Posted by KumaKuma View Post
    Sorry...im poor in calculus..not very clear with your provided answer, can you explain further??
    There's no calculus involved here.

    Do you understand how to use the basic comparison test?
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    There's no calculus involved here.

    Do you understand how to use the basic comparison test?
    It is your best friend when proving the convergence of a series.
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  6. #6
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    I did some searchings and readings about basic comparison test but i dont fully understand them.
    Is the comparison between arctan k/1+kČ with Pi/4 and with (Pi/2)/(1/kČ) necessary?
    Can i just compare in this manner?
    0 < arctan k/ (1+kČ) < 2/kČ
    I could not understand why that the numerator in the right equation has to be 2? can we use arctan k to replace the 2?
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  7. #7
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    That was slightly confusing. I'll try to put them into proper notation.
    What i meant was can i do the comparison without comparing \frac{\pi}{4} and \left( \frac{\pi}{2} \right) \, \frac{1}{k^2} ?
    And can i compare it like this:
    0 < \frac{\arctan k}{1 + k^2} < \frac{\arctan k}{k^2} instead of using <\frac{2}{k^2} ? Why the numerator is chosen to be 2?
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  8. #8
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    Hello,

    The factor is 2 because pi/2 is slightly inferior to 2. Actually, it's not a problem, we don't care about the constant factor. All we want to know is if we can compare the series with a Riemann series, which is the case, with a constant factor (not influencing the convergence) =)

    arctan is always < pi/2
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  9. #9
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    Ohh that cleared my doubts
    So it will be like:
    Since \arctan k \leq \frac{\pi}{2}

    0 < \frac{\arctan k}{1 + k^2} < \frac{2}{k^2}

    \sum_{k=1}^{\infty} \frac{2}{k^2} = 2\sum_{k=1}^{\infty} \frac{1}{k^2}

    Since \frac{2}{k^2} converges, and thus \frac{\arctan k}{1 + k^2} converges according to the basic comparison theorem.

    Am I getting it right?
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  10. #10
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    Quote Originally Posted by KumaKuma View Post
    Ohh that cleared my doubts
    So it will be like:
    Since \arctan k \leq \frac{\pi}{2}

    0 < \frac{\arctan k}{1 + k^2} < \frac{2}{k^2}

    \sum_{k=1}^{\infty} \frac{2}{k^2} = 2\sum_{k=1}^{\infty} \frac{1}{k^2}

    Since \frac{2}{k^2} converges, and thus \frac{\arctan k}{1 + k^2} converges according to the basic comparison theorem.

    Am I getting it right?
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  11. #11
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    Ah a handclap, finally
    Thanks to all of you who have helped me!
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