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Math Help - Series convergance?

  1. #1
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    Series convergance?

    does these series conv. or diver.?

    1. to infiity n=1 (arctan(n))/(n^3)
    im not sure where to start exactly, i was thinking Ratio or root test but im lost

    2. to infinity n=1  ((-3)^n-1) /(4^n)<br />
    same as above, not sure


    any help/tips would be greatly appreciated

    Thanks
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  2. #2
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    1. arctan(n)<pi/2

    So use comparison test with the conv pseries 1/n^3

    2. rewrite (-3/4)^n - (1/4)^n the sum of 2 convergent geometric series
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  3. #3
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    Quote Originally Posted by Itsygo View Post
    does these series conv. or diver.?

    1. to infiity n=1 (arctan(n))/(n^3)
    im not sure where to start exactly, i was thinking Ratio or root test but im lost

    2. to infinity n=1  ((-3)^n-1) /(4^n)<br />
    same as above, not sure


    any help/tips would be greatly appreciated

    Thanks
    I'll have a think about these, but for 1., I would say you could use a comparison test (could someone verify?)

    Since -\frac{\pi}{2}\leq\arctan{n}\leq\frac{\pi}{2} we can say that

    \sum_{n = 1}^{\infty}{\frac{\arctan{n}}{n^3}} \leq \frac{\pi}{2}\sum_{n = 1}^{\infty}{\frac{1}{n^3}}, which is a p-series.

    This p-series converges to a number, and since the original series is less than this number, the original series must also converge.
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