## Calculus: Early Transcendentals 8th Edition

Use the Ratio Test:$\lim\limits_{n \to \infty}\Big|\frac{a_{n+1}}{a_n}\Big|$. Use the recursive definition $a_{n+1}=\frac{5n+1}{4n+3}a_n$. Using substitution, $\lim\limits_{n \to \infty}\Big|\frac{\frac{5n+1}{4n+3}a_n}{a_n}\Big|$. The $a_n$ terms cancel out each other, and we are left with $\lim\limits_{n \to \infty}\Big|\frac{5n+1}{4n+3}\Big|=\frac{5}{4}>1$. Since this limit is greater than one, $\Sigma a_n$ is divergent.