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Math Help - is this correct? (convergence)

  1. #1
    Junior Member
    Joined
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    is this correct? (convergence)

    determine the convergence or divergence by any method

    infinite
    sigma 1/((n^2)+1)^1/2
    n=1

    then i used trigonometric substitution, because n^2 + 1 = a^2 + x^2, so..

    n = tan(Θ), dn = sec^2 (Θ) dΘ = tan^2 (Θ) + 1

    lim: integral (r, 1): dn/((n^2)+1)^1/2
    r->infinite

    =

    lim: integral (r, 1): [sec^2 (Θ)dΘ]/[tan^2 (Θ) + 1]^1/2
    r->infinite

    =

    lim: integral (r, 1): [sec^2 (Θ)dΘ]/sec(Θ)dΘ
    r-> infinite

    =

    lim: integral (r, 1): sec(Θ)dΘ
    r-> infinite

    lim ln|sec(r) + tan(r)| - ln|sec(1) + tan(1)| = infinite
    r->infinite

    does not converge
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    I thought it was a series, but, we can just bound the integrand for n\ge1, then \frac{1}{\sqrt{n^{2}+1}}>\frac{1}{\sqrt{n^{2}+3n^{  2}}}=\frac{1}{2n}, thus this diverges.
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