1. ## 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?

2. Originally Posted by Kuma
series: Sigma n from 1 to infinity arctan(n) / n^3/2 I want to determine convergence/divergence.
$\dfrac{{\arctan (n)}}
{{\sqrt[3]{{n^2 }}}} \leqslant \dfrac{{\frac{\pi }
{2}}}
{{\sqrt[3]{{n^2 }}}}$