## Calculus: Early Transcendentals 8th Edition

Given: $\Sigma_{n=1}^{\infty}\frac{cos(\pi n)}{\sqrt n}$ The integral test requires a function that is positive and decreasing on $[1,\infty)$ . The $cos \pi n$ alternates between $-1$ and $+1$ , so it is not positive and decreasing for the entire interval. Thus the integral test can not be applied.