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.
$\sum_{k = 1}^{\infty} \frac{k}{{(1 + k^{2})}^{p}}$

$The integral \int_{x = 1}^{\infty} \frac{x}{{(1 + x^{2})}^{p}} dx converges for p > 1 . Thus, the series \sum_{k = 1}^{\infty} \frac{k}{{(1 + k^{2})}^{p}} is convergent for p > 1 .$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given information is:

$\sum_{k = 1}^{\infty} \frac{k}{{(1 + k^{2})}^{p}}$

Step 2. Examining nature of given function:

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

Step 3. Solving the integral:

$Consider the integral : \int_{x = 1}^{\infty} f (x) dx = \int_{x = 1}^{\infty} \frac{x}{{(1 + x^{2})}^{p}} dx . Therefore, \int_{x = 1}^{\infty} f (x) dx = \lim_{k \to \infty} \int_{x = 1}^{k} \frac{x}{{(1 + x^{2})}^{p}} dx = \frac{1}{2} \lim_{k \to \infty} \int_{u = 2}^{k^{2} + 1} \frac{du}{u^{p}} (Put 1 + x^{2} = u, \Rightarrow 2 xdx = du) = \frac{1}{2} \lim_{k \to \infty} {[\frac{u^{- p + 1}}{- p + 1}]}_{2}^{k^{2} + 1} (Integrating) = \frac{1}{2 (- p + 1)} \lim_{k \to \infty} {[\frac{1}{u^{p - 1}}]}_{2}^{k^{2} + 1} (Substitution)$

Step 4. Result:

$The improper integral converges to finite value only when p > 1 . Therefore, the integral \int_{x = 1}^{\infty} \frac{x}{{(1 + x^{2})}^{p}} dx converges for p > 1 . Thus, the series \sum_{k = 1}^{\infty} \frac{k}{{(1 + k^{2})}^{p}} is convergent for p > 1 .$

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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.
$\sum_{k = 1}^{\infty} \frac{k}{{(1 + k^{2})}^{p}}$

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∞k(1+k2)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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In Exercises 48–51 find all values of p so that the series converges.
$\sum_{k = 1}^{\infty} \frac{k}{{(1 + k^{2})}^{p}}$