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 10.2 - Infinite Series - Exercises 10.2 - Page 580: 64

Answer

Diverges

Work Step by Step

Formula to calculate the sum of a geometric series is: $S=\dfrac{a}{1-r}$; Since, we have two series$\sum_{n =1}^{ \infty}\dfrac{2^n+4^n}{3^n+4^n}=\dfrac{\sum_{n =1}^{ \infty}(\dfrac{1}{2})^n+1}{\sum_{n =1}^{ \infty}(\dfrac{3}{4})^n+1}= 1\ne 0$ Hence, $\sum_{n =1}^{ \infty}\dfrac{2^n+4^n}{3^n+4^n}$ diverges as per Nth Term Integral test.
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