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: 81

Answer

See the picture below. The sequence converges to 0.5.
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Work Step by Step

If we write the first 10 terms, we can make a conjucture of the convergence of the sequence. $a_n=\frac{n+4}{2n}$ $a_1=\frac{1+4}{2\times 1}=2.5$ $a_2=\frac{2+4}{2\times 2}=1.5$ $a_3=\frac{3+4}{2\times 3}\approx 1.17$ $a_4=\frac{4+4}{2\times 4}=1$ $a_5=\frac{5+4}{2\times 5}=0.9$ $a_6=\frac{6+4}{2\times 6}\approx 0.84$ $a_7=\frac{7+4}{2\times 7}\approx 0.79$ $a_8=\frac{8+4}{2\times 8}=0.75$ $a_9=\frac{9+4}{2\times 9}\approx 0.72$ $a_{10}=\frac{10+4}{2\times 10}=0.7$ Also, if we look at really huge numbers, for example 10,000 and 100,000 we can see the convergence. $a_{10000}=\frac{10000+4}{2\times 10000}=0.5002$ $a_{100000}=\frac{100000+4}{2\times 100000}=0.50002$
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