Problem 48
Question
Use the Direct Comparison Test to determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \frac{1}{4 \sqrt[3]{n}-1} $$
Step-by-Step Solution
Verified Answer
By Direct Comparison Test, the series \(\sum_{n=1}^{\infty} \frac{1}{4 \sqrt[3]{n}-1}\) is divergent.
1Step 1 - Identify a Comparable Series
Since \(\sqrt[3]{n}\) is the dominant term in the denominator, the series can be compared with \(\sum_{n=1}^{\infty} \frac{1}{\sqrt[3]{n}}\).
2Step 2 - 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.
Other exercises in this chapter
Problem 48
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