Answer
Convergent
Work Step by Step
$\lim\limits_{x \to \infty} \int_1^\infty \sec h^2 x dx=\lim\limits_{k \to \infty} [\tan h x]_{1}^{k}$
and $ \lim\limits_{k \to \infty}[\tanh (k)-\tanh (1)]=(1)- \tanh (1)$
so, $ (1)- \tanh (1) \approx 0.76$
Thus, the series is Convergent