## Thomas' Calculus 13th Edition

Let us consider $a_n=\dfrac{1}{ n^{3/2}}$ and $u_n$ refers to all positive values of $n$. Here, $f(n)=\dfrac{1}{ n^{3/2}}$; $f'(n)=\dfrac{-3}{2\sqrt {n^5}} \lt 0$ The negative sign simplifies that the sequence $u_n$ is not increasing. Then, $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty}\dfrac{1}{ n^{(3/2)}}=0$ Hence, the series converges by the Alternating Series Test.