## Thomas' Calculus 13th Edition

Let $u_n=\Sigma_{n=1}^\infty\dfrac{\sec^{-1} n}{n^{(1.3)}}$ We compare with: $\Sigma_{n=1}^\infty\dfrac{\sec^{-1} n}{n^{(1.3)}} \lt \Sigma_{n=1}^\infty \dfrac{{\dfrac{\pi}{2}}}{n^{(1.3)}}$ Here, $\Sigma_{n=1}^\infty \dfrac{1}{n^{(1.3)}}$ shows a convergent $p$-series with $p=1.3 \gt 1$ Thus, the series converges by the direct comparison test.