Answer
Convergent
Work Step by Step
Since, we have:
$\int_1^\infty\sec h x dx=2\lim\limits_{a \to \infty} \int_1^a\frac{e^x}{1+(e^x)^2} dx$
or, $=2 [\lim\limits_{a \to \infty} [tan^{-1}(e^x)]|_{1}^{a}$
or, $ =2 \lim\limits_{a \to \infty}(\tan^{-1} e^a-tan^{-1} e)$
or, $=\pi-2 tan^{-1} e$
or, $\approx 0.71$
Hence, the series is Convergent
