## Calculus: Early Transcendentals 8th Edition

Ratio Test: $\lim\limits_{n \to \infty}\frac{a_{n+1}}{a_{n}}=\lim\limits_{n \to \infty}\frac{\frac{(n+1)^{3}}{5^{n+1}}}{\frac {n^{3}}{5^{n}}}$ $=\frac{1}{5}(\frac{n+1}{n})^{3}$ $=\frac{1}{5}\lt 1$ Thus, the series is convergent.