Chapter 7: Q. 27 (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.
$\sum_{k = 2}^{\infty} \frac{1}{k \sqrt{\ln k}}$

Short Answer

Expert verified

Ans: The series $\sum_{k = 2}^{\infty} \frac{1}{k \sqrt{\ln k}}$ is divergent.

Step by step solution

Step 1. Given information.

given,

$\sum_{k = 2}^{\infty} \frac{1}{k \sqrt{\ln k}}$

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 $f (x) = \frac{1}{x \sqrt{\ln x}}$ .

The function $f (x) = \frac{1}{x \sqrt{\ln x}}$ 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=2∞ f(x)dx=∫x=2∞ 1xln⁡xdx

Therefore,

$\begin{aligned} \int_{x = 2}^{\infty} f (x) d x = lim_{k \to \infty} \int_{x = 2}^{k} \frac{1}{x \sqrt{\ln x}} d x \\ = lim_{k \to \infty} \int_{u = \ln 2}^{\ln k} \frac{1}{\sqrt{u}} d u (Put \ln x = u, \Rightarrow \frac{1}{x} d x = d u) \\ = lim_{k \to \infty} [2 \sqrt{u}]_{\ln 2}^{\ln k} \\ = 2 lim_{k \to \infty} [\sqrt{\ln k} - \ln 2] (Substitution) \\ = 2 (\infty - \ln 2) (Take limit) \\ = \infty \end{aligned}$

Step 4. Thus, the value of the integral is ∫x=2∞ 1xln⁡xdx=∞

The integral converges. Therefore, the series $\sum_{k = 2}^{\infty} \frac{1}{k \sqrt{\ln k}}$ is divergent.

Hence, by integral test, the series $\sum_{k = 2}^{\infty} \frac{1}{k \sqrt{\ln k}}$ is divergent

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

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.
$\sum_{k = 2}^{\infty} \frac{1}{k \sqrt{\ln k}}$

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=2∞ f(x)dx=∫x=2∞ 1xln⁡xdx

Step 4. Thus, the value of the integral is ∫x=2∞ 1xln⁡xdx=∞

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Discrete Mathematics

Probability and Statistics

Logic and Functions

Pure Maths

Decision Maths

Mechanics Maths

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. ∑k=2∞ 1kln⁡k

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=2∞ f(x)dx=∫x=2∞ 1xln⁡xdx

Step 4. Thus, the value of the integral is ∫x=2∞ 1xln⁡xdx=∞

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Discrete Mathematics

Probability and Statistics

Logic and Functions

Pure Maths

Decision Maths

Mechanics Maths

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.
$\sum_{k = 2}^{\infty} \frac{1}{k \sqrt{\ln k}}$