   Chapter 11.1, Problem 57E

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.)57. a n = ( − 1 ) n n n + 1

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=(1)nnn+1 .

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 nn+1 an=(−1)nnn+1 1 12=0.5000 −0.5000 2 23=0.6667 0.6667 3 34=0.7500 −0.7500 4 45=0.8000 0.8000 5 56=0.8333 −0.8333 6 67=0.8571 0.8571 7 78=0.8750 −0.8750 8 89=0.8889 0.8889 9 910=0.9000 −0.9000 10 1011=0.9091 0.9091 11 1112=0.9167 −0.9167 12 1213=0.9231 0.9231 13 1314=0.9286 −0.9286 14 1415=0.9333 0.9333 15 1516=0.9375 −0.9375 16 1617=0.9412 0.9412 17 1718=0.9444 −0.9444 18 1819=0.9474 0.9474 19 1920=0.9500 −0.9500 20 2021=0.9524 0.9524

Plot the points (n,an), for n=1,2,...20 on the graph as shown below in Figure 1.

From the graph, it is observed that the sequence is divergent because the sequence oscillates between −1 and 1.

To prove:

The sequence an=(1)nnn+1 is divergent.

Proof:

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

That is, compute limnan=limn((1)nnn+1) .

Use the subsequence n=2m in equation (1).

limn((1)nnn+1)=limm((1)2m2m2m+1)=limm(((1)2)m2m2m+1)=limm(1m2m2m+1)=limm(2m2m+1)

Divide the numerator and denominator by the highest power.

limm((1)2m2m2m+1)=limm(2mm2m+1m)                                =limm(22+1m)                                =limm(2)limm(2+1m)=limm(2)limm(2)+limm(1m)

Apply infinity property limn(kna)=0 and simplify the terms

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