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Q. 27

Expert-verified

Calculus

Found in: Page 625

Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

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.

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

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

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

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

.

Most popular questions for Math Textbooks

For each series in Exercises 44–47, do each of the following:

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(b) Use the 10th term in the sequence of partial sums to approximate the sum of the series.

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(d) Use your answers from parts (b) and (c) to find an interval containing the sum of the series.

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