University Calculus: Early Transcendentals (3rd Edition)

Consider $u_n=(-1)^{n+1}\dfrac{n}{n^3+1}$ Here, $u_n$ is positive for all the values of $n$. and $|u_n|=\dfrac{n}{n^3+1}$ We see that $\Sigma_{n=1}^\infty\dfrac{n}{n^3+1} \leq \Sigma_{n=1}^\infty\dfrac{n}{n^3}=\Sigma_{n=1}^\infty\dfrac{1}{n^2}$ (a convergent p-series) Thus, the series converges absolutely.