Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

In Exercises 48–51 find all values of p so that the series converges.
$\sum_{k = 1}^{\infty} \frac{\ln (k)}{k^{p}}$

$The integral \int_{x = 1}^{\infty} \frac{\ln (x)}{x^{p}} dx converges for p > 1 . Thus, the series \sum_{k = 1}^{\infty} \frac{\ln (k)}{k^{p}} is convergent for p > 1 .$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given information is:

$\sum_{k = 1}^{\infty} \frac{\ln (k)}{k^{p}}$

Step 2. Examining nature of given function:

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

Step 3. Solving the integral:

$Consider the integral : \int_{x = 1}^{\infty} f (x) dx = \int_{x = 1}^{\infty} \frac{\ln (x)}{x^{p}} dx . Therefore, \int_{x = 1}^{\infty} f (x) dx = \lim_{k \to \infty} \int_{x = 1}^{k} \frac{\ln (x)}{x^{p}} dx = \lim_{k \to \infty} \int_{u = 0}^{\ln (k)} u {(e^{u})}^{1 - p} du (Put \ln (x) = u, \Rightarrow \frac{1}{x} dx = du)$

Step 4. Result:

$The improper integral converges to finite value only when p > 1 . Therefore, the integral \int_{x = 1}^{\infty} \frac{\ln (x)}{x^{p}} dx converges for p > 1 . Thus, the series \sum_{k = 1}^{\infty} \frac{\ln (k)}{k^{p}} is convergent for p > 1 .$

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Calculus

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Short Answer

In Exercises 48–51 find all values of p so that the series converges.
$\sum_{k = 1}^{\infty} \frac{\ln (k)}{k^{p}}$

Step by Step Solution

Step 1. Given information is:

Step 2. Examining nature of given function:

Step 3. Solving the integral:

Step 4. Result:

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Calculus

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Short Answer

In Exercises 48–51 find all values of p so that the series converges.∑k=1∞lnkkp

Step by Step Solution

Step 1. Given information is:

Step 2. Examining nature of given function:

Step 3. Solving the integral:

Step 4. Result:

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In Exercises 48–51 find all values of p so that the series converges.
$\sum_{k = 1}^{\infty} \frac{\ln (k)}{k^{p}}$