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Q. 21

Expert-verified

Calculus

Found in: Page 692

Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

Use an appropriate Maclaurin series to find the values of the series in Exercises 17–22.
$\sum_{k = 0}^{\infty} \frac{{(- 1)}^{k} π^{2 k + 1}}{(2 k + 1)!}$

The required answer is $\sum_{k = 0}^{\infty} \frac{{(- 1)}^{k} π^{2 k + 1}}{(2 k + 1)!} = 0$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given Information

The given series is $\sum_{k = 0}^{\infty} \frac{{(- 1)}^{k} π^{2 k + 1}}{(2 k + 1)!}$

Step 2. Explanation

The maclaurin series for the function $f (x) = \sin x is \sin x = \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k} x^{2 k + 1}}{(2 k + 1)!}$

So, the given series is the maclaurin series for $\sin x a t x = π$

Since, $\sum_{k = 0}^{\infty} \frac{{(- 1)}^{k} x^{2 k + 1}}{(2 k + 1)!} = \sin x$

Thus,

$\sum_{k = 0}^{\infty} \frac{{(- 1)}^{k} π^{2 k + 1}}{(2 k + 1)!} = sinπ \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k} π^{2 k + 1}}{(2 k + 1)!} = 0$

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