## Thomas' Calculus 13th Edition

Let us consider $a_n=\dfrac{1}{\sqrt n}$ Here, we have $f(n)=\dfrac{1}{\sqrt n} \implies f'(n)=\dfrac{-1}{2\sqrt {n^3}} \lt 0$ The negative sign shows that, the sequence $u_n$ is decreasing. Then, $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty}\dfrac{1}{\sqrt n}=\lim\limits_{n \to \infty}\dfrac{1}{ n^{1/2}}=0$ So, the series converges by the Alternating Series Test. Here, the series $\Sigma_{n=1}^\infty \dfrac{1}{n^{1/2}}$ shows a $p$-series having $p=\dfrac{1}{2}$, thus the series diverges Therefore, the series do not converge absolutely this implies that the series is conditionally convergent.