Answer
Convergent
Work Step by Step
$\lim\limits_{x \to \infty} \int_1^\infty\sec x dx=(2) \lim\limits_{k \to \infty} [\dfrac{e^x}{1+e^x}]_{1}^{k}$
and $(2) \lim\limits_{k \to \infty}(\tan^{-1} e^{(k)}-\tan^{-1} e^{(1)})=(\pi)-2 \tan^{-1} e$
so, $(\pi)-(2) \tan^{-1} (e) \approx 0.71$
Thus, the series is Convergent.