## Calculus: Early Transcendentals 8th Edition

Given: $\Sigma_{n=1 }^{\infty}\frac{n\pi^{n}}{(-3)^{n-1}}$ Re-write the given series. $\Sigma_{n=1 }^{\infty}\frac{n\pi^{n}}{(-3)^{n-1}}=\Sigma_{n=1 }^{\infty}(-1)^{n}3n(\pi/3)^{n}$ $L=\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}\frac{(-1)^{n+1}3(n+1)(\pi/3)^{n+1} }{(-1)^{n}3n(\pi/3)^{n}}$ $=\lim\limits_{n \to \infty}\frac{n+1}{n}\frac{\pi}{3}$ $=\lim\limits_{n \to \infty}[1+\frac{1}{n}]\frac{\pi}{3}$ $=[1+0]\frac{\pi}{3}$ $=\frac{\pi}{3}\gt 1$ Hence, the series diverges by ratio test.