23–38. Divergence, Integral, and p-series Tests Use the Divergence Test, the Integral Test, or the p-series test to determine whether the following series converge.
∑ (k = 1 to ∞) 1 / ∛k

Verified step by step guidance

1. Identify the given series: \( \sum_{k=1}^{\infty} \frac{1}{\sqrt[3]{k}} \). This can be rewritten as \( \sum_{k=1}^{\infty} \frac{1}{k^{1/3}} \).
2. Recognize that this is a p-series of the form \( \sum_{k=1}^{\infty} \frac{1}{k^p} \) where \( p = \frac{1}{3} \).
3. Recall the p-series test: A p-series \( \sum \frac{1}{k^p} \) converges if and only if \( p > 1 \), and diverges otherwise.
4. Since \( p = \frac{1}{3} < 1 \), the p-series test indicates that the series diverges.
5. Optionally, you could confirm this result using the Integral Test by evaluating the improper integral \( \int_1^{\infty} \frac{1}{x^{1/3}} \, dx \) and checking if it converges or diverges.

Key Concepts

Divergence Test

The Divergence Test states that if the limit of the terms of a series does not approach zero as k approaches infinity, the series diverges. It is a quick initial check but cannot confirm convergence if the limit is zero.

Integral Test

The Integral Test compares a series to an improper integral of a related continuous, positive, decreasing function. If the integral converges, the series converges; if the integral diverges, so does the series.

p-series Test

A p-series is of the form ∑ 1/k^p. It converges if p > 1 and diverges if p ≤ 1. This test helps quickly determine convergence for series with terms involving powers of k.