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Math Help - 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.
    \dfrac{{\arctan (n)}}<br />
{{\sqrt[3]{{n^2 }}}} \leqslant \dfrac{{\frac{\pi }<br />
{2}}}<br />
{{\sqrt[3]{{n^2 }}}}
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