College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 7 - Section 7.1 - Sequences and Series - 7.1 Exercises - Page 636: 82

Answer

See the picture below. The sequence converges to 2.

Work Step by Step

If we write the first 10 terms, we can make a conjucture of the convergence of the sequence. $a_n=\frac{1+4n}{2n}$ $a_1=\frac{1+4\times 1}{2\times 1}=2.5$ $a_2=\frac{1+4\times 2}{2\times 2}=2.25$ $a_3=\frac{1+4\times 3}{2\times 3}\approx 2.17$ $a_4=\frac{1+4\times 4}{2\times 4}\approx 2.13$ $a_5=\frac{1+4\times 5}{2\times 5}\approx 2.1$ $a_6=\frac{1+4\times 6}{2\times 6}\approx 2.08$ $a_7=\frac{1+4\times 7}{2\times 7}\approx 2.07$ $a_8=\frac{1+4\times 8}{2\times 8}\approx 2.06$ $a_9=\frac{1+4\times 9}{2\times 9}\approx 2.055$ $a_{10}=\frac{1+4\times 10}{2\times 10}\approx 2.05$ Also, if we look at really huge numbers, for example 10,000 and 100,000 we can see the convergence. $a_{10000}=\frac{1+4\times 10000}{2\times 10000}\approx 2.00005$ $a_{100000}=\frac{1+4\times 100000}{2\times 100000}=2.000005$
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