Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.7 - Power Series - Exercises 10.7 - Page 613: 47

Answer

Interval of Convergence: $ -\sqrt 2 \lt x \sqrt 2$ and $\Sigma_{n=0}^{\infty} (\dfrac{x^2+1}{3})^n=\dfrac{3}{2-x^2}$

Work Step by Step

Remember that the series will converge when $|f(x)|=|\dfrac{x^2+1}{3}| \lt 1$ This implies that $x^2+1 \lt 3$ and $x^2 \lt 2$ This means that the interval of convergence is: $ -\sqrt 2 \lt x \sqrt 2$ Next, we will find the sum of the given geometric series $S=\dfrac{a}{1-r} \\=\dfrac{1}{1-\dfrac{x^2+1}{3}} \\=\dfrac{3}{2-x^2}$ So, $\Sigma_{n=0}^{\infty} (\dfrac{x^2+1}{3})^n=\dfrac{3}{2-x^2}$
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