University Calculus: Early Transcendentals (3rd Edition)

Since, we have $\Sigma_{n=1}^\infty\dfrac{\sec^{-1} n}{n^{1.3}}$ and $\Sigma_{n=1}^\infty\dfrac{\sec^{-1} n}{n^{1.3}} \lt \Sigma_{n=1}^\infty \dfrac{\pi/2}{n^{1.3}}$ This is a p-series with $p=1.3 \gt 1$ and is convergent. Hence, the given series converges by the direct comparison test.