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