Question

Use the Direct Comparison Test to determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \frac{1}{4 \sqrt[3]{n}-1} $$

Short answer

By Direct Comparison Test, the series \(\sum_{n=1}^{\infty} \frac{1}{4 \sqrt[3]{n}-1}\) is divergent.

Step-by-step solution

- Identify a Comparable Series

Since \(\sqrt[3]{n}\) is the dominant term in the denominator, the series can be compared with \(\sum_{n=1}^{\infty} \frac{1}{\sqrt[3]{n}}\).

- Apply Direct Comparison Test

The series \(\sum_{n=1}^{\infty} \frac{1}{\sqrt[3]{n}}\) is a p-series where p = 1/3 which is less than 1. Therefore, it is divergent. For all n, \(\frac{1}{4 \sqrt[3]{n}-1}\) is less than or equal to \( \frac{1}{\sqrt[3]{n}}\). So by the Direct Comparison Test, the series \(\sum_{n=1}^{\infty} \frac{1}{4 \sqrt[3]{n}-1}\) is also divergent.