## Calculus 8th Edition

Using the comparison test, we know the series on the left side is always greater than the series on the right side because the numerator on the left if always larger, and the denominator is always smaller. The Comparison Test states that the p-series $\sum_{n=1}^{\infty}\frac{1}{n^{p}}$ is convergent if $p\gt 1$ and divergent if $p\leq 1$. $\Sigma _{n=1}^{\infty}\frac{4^{n+1}}{3^{n}-2)^{4}} \geq \Sigma _{n=1}^{\infty}\frac{4^{n}}{3^{n}}$ Re-arranging the exponents. $\Sigma _{n=1}^{\infty}\frac{4^{n+1}}{3^{n}-2)^{4}} \geq \Sigma _{n=1}^{\infty}(\frac{4}{3})^{n}$ We find that the series on the right side is a geometric-series with $|r|\gt 1$ , so the series on the right side diverges and therefore, the series on the left side diverges also.