Chapter 9 - Section 9.5 - Absolute Convergence; The Ratio and Root Tests - Exercises - Page 516: 24

Converges

Consider $a_n=\dfrac{(-2)^n}{3^n}$ Here, $\Sigma_{n=1}^\infty \dfrac{(-2)^n}{3^n} \leq \Sigma_{n=1}^\infty \dfrac{2^n}{3^n}=\Sigma_{n=1}^\infty (\dfrac{2}{3})^{n} $ we see that $\Sigma_{n=1}^\infty (\dfrac{2}{3})^{n}$ is a geometric convergent series. Hence, the given series converges by the direct comparison test.