# How do you use the ratio test to test the convergence of the series ∑ 3^n/(4n³+5) from n=1 to infinity?

Oct 18, 2015

The series is divergent, see the explanation.

#### Explanation:

$L = {\lim}_{n \to \infty} | {a}_{n + 1} / {a}_{n} |$

$L = {\lim}_{n \to \infty} | \frac{{3}^{n + 1} / \left(4 {\left(n + 1\right)}^{3} + 5\right)}{{3}^{n} / \left(4 {n}^{3} + 5\right)} | = {\lim}_{n \to \infty} | \frac{\frac{{3}^{n} \cdot 3}{4 {\left(n + 1\right)}^{3} + 5}}{{3}^{n} / \left(4 {n}^{3} + 5\right)} |$

$L = {\lim}_{n \to \infty} \frac{3 \left(4 {n}^{3} + 5\right)}{4 {\left(n + 1\right)}^{3} + 5} = {\lim}_{n \to \infty} \frac{12 {n}^{3} + 15}{4 \left({n}^{3} + 3 {n}^{2} + 3 n + 1\right) + 5}$

$L = {\lim}_{n \to \infty} \frac{12 {n}^{3} + 15}{4 {n}^{3} + 12 {n}^{2} + 12 n + 9} = {\lim}_{n \to \infty} \frac{12 + \frac{15}{n} ^ 3}{4 + \frac{12}{n} + \frac{12}{n} ^ 2 + \frac{9}{n} ^ 3}$

$L = \frac{12}{4} = 3 > 1 \implies$ the series is divergent