   Chapter 11.1, Problem 20E

Chapter
Section
Textbook Problem

# Calculate, to four decimal places, the first ten terms of the sequence and use them to plot the graph of the sequence by hand. Does the sequence appear to have a limit? If so, calculate it. If not, explain why.20. a n = 2 + ( − 1 ) n n

To determine

To find: The first ten terms of the sequence and plot them on the graph to obtain the limit.

Explanation

Given:

The sequence is an=2+(1)nn . (1)

Theorem used:

If limn|an|=0 , then limnan=0 . (2)

Calculation:

Obtain the first ten terms of the sequence.

Substitute 1 for n in equation (1).

a1=2+(1)11   =21   =1

Thus, the first term of the sequence is a1=1.0000_ .

Substitute 1 for n in equation (2).

a2=2+(1)22   =2+12   =52   =2.5

Thus, the second term of the sequence is a2=2.5000_ .

Substitute 3 for n in equation (1).

a3=2+(1)33=213=53=1.6667

Thus, the third term of the sequence is a3=1.6667_ .

Substitute 4 for n in equation (1).

a4=2+(1)44   =2+14   =94   =2.25

Thus, the fourth term of the sequence is a4=2.2500_ .

Substitute 5 for n in equation (1).

a5=2+(1)55   =215   =95   =1.8

Thus, the fifth term of the sequence is a5=1.8000_ .

Substitute 6 for n in equation (1).

a6=2+(1)66   =2+16   =136   =2.1667

Thus, the sixth term of the sequence is a6=2.1667_ .

Substitute 7 for n in equation (1).

a7=2+(1)77   =217   =137   =1.8571

Thus, the seventh term of the sequence is a7=1.8571_ .

Substitute 8 for n in equation (10.

a8=2+(1)88   =2+18   =178   =2.125

Thus, the eighth term of the sequence is a8=2.1250_ .

Substitute 9 for n in equation (1).

a9=2+(1)99   =219   =179   =1.8889

Thus, the ninth term of the sequence is a9=1

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