I thought it was a series, but, we can just bound the integrand for then thus this diverges.
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