   Chapter 11.1, Problem 63E

Chapter
Section
Textbook Problem

# Use a graph of the sequence to decide whether the sequence is convergent or divergent. If the sequence is convergent, guess the value of the limit from the graph and then prove your guess. (See the margin note on page 739 for advice on graphing sequences.)63. a n = 1 ⋅ 3 ⋅ 5 ⋅ ⋯ ( 2 n − 1 ( 2 n ) n

To determine

Whether the sequence is convergent or divergent; guess the value of the limit if convergent with valid proof.

Explanation

Given:

The sequence is an=135...(2n1)(2n)n .

Definition used:

If an is a sequence and limnan exists, then the sequence an is said to be convergent; otherwise it is divergent.

Calculation:

Obtain the first 20 terms of the sequence.

 n 1⋅3⋅5⋅...⋅(2n−1) (2n)n an=1⋅3⋅5⋅...⋅(2n−1)(2n)n 1 1 21=2 121=0.5000 2 3 42=16 342=0.1875 3 15 63=216 1563=0.0694 4 105 84=4,096 10584=0.0256 5 945 105=100,000 945105=0.0095 6 10,395 126=2,985,984 10,395126=0.0104 7 135,135 147=105,413,504 135,135147=0.0013 8 2,027,025 168=4,294,967,296 2,027,025168=0.0005 9 34,459,425 189=198,359,290,368 34,459,4259!=0.0002 10 654,729,075 2010=10,240,000,000,000 654,729,0752010=0.0001

Plot the points (n,an), for n=1,2,...10 on the graph as shown below in Figure 1.

From the graph, it is observed that the sequence is convergent.

Also, the value of the limit is guessed to 0 as the plotted points are closer to 0.

To prove:

The sequence an=135...(2n1)(2n)n is convergent to 0.

Proof:

Obtain the limit of the sequence (the value of the term an as n tends to infinity)

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