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Thread: series convergence

  1. #1
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    series convergence

    Hi.

    series: Sigma n from 1 to infinity arctan(n) / n^3/2

    I want to determine convergence/divergence.

    What i tried:

    f(x) = arctan(x) / x^3/2

    arctan(x) > f(x)

    since arctan(x) converges to pi/2 as x -> inf, f(x) should converge too.
    Is that reasoning right?
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  2. #2
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    Quote Originally Posted by Kuma View Post
    series: Sigma n from 1 to infinity arctan(n) / n^3/2 I want to determine convergence/divergence.
    $\displaystyle \dfrac{{\arctan (n)}}
    {{\sqrt[3]{{n^2 }}}} \leqslant \dfrac{{\frac{\pi }
    {2}}}
    {{\sqrt[3]{{n^2 }}}}$
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