Chapter 7: Q. 30 (page 625)

Explain why the integral test may be used to analyze the given series and then use the test to determine whether the series converges or diverges.
$\sum_{k = 1}^{\infty} (3 k - 1)^{- 2 / 3}$

Short Answer

Expert verified

Ans: The series $\sum_{k = 1}^{\infty} (3 k - 1)^{- 2 / 3}$ is divergent.

Step by step solution

Step 1. Given information.

given,

$\sum_{k = 1}^{\infty} (3 k - 1)^{- 2 / 3}$

Step 2. The objective is to explain why the integral test is used to determine the convergence or divergence of the series and use the test to determine the convergence or divergence of the series.

Consider function $f (x) = (3 x - 1)^{- \frac{2}{3}}$ .

The function $f (x) = (3 x - 1)^{- \frac{2}{3}}$ is continuous, decreasing, with positive terms. Therefore, all the conditions of the integral test are fulfilled. So, the integral test is applicable.

Step 3. Consider the integral ∫x=1∞ f(x)dx=∫x=1∞ (3x−1)−23dx.

Therefore,

$\begin{aligned} \int_{x = 1}^{\infty} f (x) d x = lim_{k \to \infty} \int_{x = 1}^{k} (3 x - 1)^{- \frac{2}{3}} d x \\ = \frac{1}{3} lim_{k \to \infty} \int_{u = 2}^{3 k - 1} u^{- 2 / 3} d u (Put 3 x - 1 = u \Rightarrow 3 d x = d u) \\ = lim_{k \to \infty} {[u^{1 / 3}]}_{2}^{3 k - 1} (Integrating) \\ = lim_{k \to \infty} [(3 k - 1)^{1 / 3} - 2^{1 / 3}] (Substitution) \\ = \infty \end{aligned}$

Step 4. Thus, the value of the integral is ∫x=1∞ (3x−1)−23dx=∞

The integral converges. Therefore, the series $\sum_{k = 1}^{\infty} (3 k - 1)^{- \frac{2}{3}}$ is divergent.

Hence, by integral test, the series $\sum_{k = 1}^{\infty} (3 k - 1)^{- \frac{2}{3}}$ is divergent.

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Explain why the integral test may be used to analyze the given series and then use the test to determine whether the series converges or diverges.
$\sum_{k = 1}^{\infty} (3 k - 1)^{- 2 / 3}$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. The objective is to explain why the integral test is used to determine the convergence or divergence of the series and use the test to determine the convergence or divergence of the series.

Step 3. Consider the integral ∫x=1∞ f(x)dx=∫x=1∞ (3x−1)−23dx.

Step 4. Thus, the value of the integral is ∫x=1∞ (3x−1)−23dx=∞

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Explain why the integral test may be used to analyze the given series and then use the test to determine whether the series converges or diverges. ∑k=1∞ (3k−1)−2/3

Short Answer

Step by step solution

Step 1. Given information.

Step 2. The objective is to explain why the integral test is used to determine the convergence or divergence of the series and use the test to determine the convergence or divergence of the series.

Step 3. Consider the integral ∫x=1∞ f(x)dx=∫x=1∞ (3x−1)−23dx.

Step 4. Thus, the value of the integral is ∫x=1∞ (3x−1)−23dx=∞

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

Geometry

Pure Maths

Applied Mathematics

Statistics

Discrete Mathematics

Study anywhere. Anytime. Across all devices.

Explain why the integral test may be used to analyze the given series and then use the test to determine whether the series converges or diverges.
$\sum_{k = 1}^{\infty} (3 k - 1)^{- 2 / 3}$