## Calculus 8th Edition

Given: $\Sigma_{n=1}^\infty\frac{(-1)^{n}}{\sqrt {n+1}}$ Here, $b_{n}= \frac{1}{\sqrt {n+1}}$ $b_{n}= \frac{1}{\sqrt {n+1}}$ is decreasing $\sqrt {n+1}$ is increasing. Also, $\lim\limits_{n \to \infty}b_{n}=\lim\limits_{n \to \infty}\frac{1}{\sqrt {n+1}}=\frac{1}{\sqrt {\infty+1}}=0$ Hence, the series is convergent by Alternating Series Test.