## University Calculus: Early Transcendentals (3rd Edition)

Consider $a_n=(-1)^{n+1}\dfrac{1+n}{n^2}$ $\Sigma_{n=1}^\infty |u_n|=\Sigma_{n=1}^\infty \dfrac{1+n}{n^2}=\Sigma_{n=1}^\infty \dfrac{1}{n^2}+\Sigma_{n=1}^\infty \dfrac{1}{n}$ Here, $\Sigma_{n=1}^\infty \dfrac{1}{n^2}$ a convergent p-series and $\Sigma_{n=1}^\infty \dfrac{1}{n}$ is a divergent p-series. Thus, the sum of both series is divergent. Next, consider $b_n=\dfrac{1}{n^2}+\dfrac{1}{n}$ we can see that $b_n$ is decreasing and $\lim\limits_{n \to \infty} b_n=\lim\limits_{n \to \infty} \dfrac{1}{n^2}+\lim\limits_{n \to \infty} \dfrac{1}{n}=0$ Hence, the series Converges Conditionally by the Alternating Series Test.