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 10.1 - Sequences - Exercises 10.1 - Page 572: 140

Answer

$ a.\quad$ Diverges. See image. Steps given below. $ b.\quad$ Diverges.

Work Step by Step

$a_{1}=1$ $a_{2}=1+(-2)=-1$ $a_{3}=1+(-2)+(-2)^{2}$ $a_{4}=1+(-2)+(-2)^{2}+(-2)^{3}$ $...$ $a_{n}=\displaystyle \sum_{k=1}^{n}(-2)^{n-1}$ = sum of the first n terms of a geometric sequence $a_{n}= \displaystyle \frac{1-(-2)^{k}}{1-(-2)}=\frac{1-(-2)^{n}}{3}$ $a.\quad $ The steps you take will depend on the CAS you are using, but they follow the same logic. Using the free online CAS at geogebra.org/cas: Cell 1: Enter the function representing the sequence $a(x)=\displaystyle \frac{5(1-5^{-x})}{4}$ From the dropdown menu, select "Table of values". In the dialog box for the table, set the range from 1 to 25, step 1. When we observe the graph, the points to the right increase in magnitude and alternate above/below the x-axis. The sequence seems to diverge. In the next free cell of the CAS, we find the limit when $ n\rightarrow\infty$ Here, we enter "L=Limit(a, infinity)" (without quotes) The CAS returns the limit to be " $?$ ". (There is no limit) $b.\quad $ The sequence diverges.
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