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Math Help - Test for convergence

  1. #1
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    Test for convergence

    Test for convergence-photo-6.jpg
    Did I go about this correctly?
    Since b_n is p series and p=3/2 which is greater than 1, it is convergent and so is a_n due to the comparison test.

    It makes sense to me but I need confirmation since Im not too comfortable with this material.

    Thank you!
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  2. #2
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    I don't agree with your upper bound...

    -\frac{\pi}{2} < \arctan{n} < \frac{\pi}{2} for all n, so surely

    -\frac{\frac{\pi}{2}}{n\sqrt{n}} < \frac{\arctan{n}}{n\sqrt{n}} < \frac{\frac{\pi}{2}}{n\sqrt{n}} (we can assume n\sqrt{n} > 0 since we are summing from n = 1 to \infty...


    So you will need to choose \frac{\frac{\pi}{2}}{n\sqrt{n}} as your upper bound.

    The rest of your logic looks fine.
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