James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 11.1 Sequences 699
a n
1
0 1 2 3 4
n
_1
FIGURE 8
The graph of the sequence in Example 8
is shown in Figure 9 and supports our
answer.
1
a n
0
1
n
_1
FIGURE 9
Example 7 Determine whether the sequence a n − s21d n is convergent or divergent.
SOLUTION If we write out the terms of the sequence, we obtain
h21, 1, 21, 1, 21, 1, 21, . . .j
The graph of this sequence is shown in Figure 8. Since the terms oscillate between 1
and 21 infinitely often, a n does not approach any number. Thus lim n l ` s21d n does not
exist; that is, the sequence hs21d n j is divergent.
n
s21d
Example 8 n
Evaluate lim if it exists.
n l ` n
SOLUTION We first calculate the limit of the absolute value:
Therefore, by Theorem 6,
lim
n l `
Z
s21d n
n
s21d n
lim
n l ` n
1
Z − lim
n l ` n − 0
The following theorem says that if we apply a continuous function to the terms of a
convergent sequence, the result is also convergent. The proof is left as Exercise 88.
− 0
7 Theorem If lim
n l `
a n − L and the function f is continuous at L, then
lim
n l `
f sa nd − f sLd
n
Creating Graphs of Sequences
Some computer algebra systems have
special commands that enable us to create
sequences and graph them directly.
With most graphing calcula tors, however,
sequences can be graphed by using
parametric equations. For instance, the
sequence in Example 10 can be graphed
by entering the parametric equations
x − t y − t!yt t
and graphing in dot mode, starting with
t − 1 and setting the t-step equal to 1.
The result is shown in Figure 10.
0
1
FIGURE 10
10
Example 9 Find lim
n l `
sinsynd.
SOLUTION Because the sine function is continuous at 0, Theorem 7 enables us to write
lim S sinsynd − sin lim
n l ` n l `
syndD − sin 0 − 0
Example 10 Discuss the convergence of the sequence a n − n!yn n , where
n! − 1 ? 2 ? 3 ? ∙ ∙ ∙ ? n.
SOLUTION Both numerator and denominator approach infinity as n l ` but here we
have no corresponding function for use with l’Hospital’s Rule (x! is not defined when x
is not an integer). Let’s write out a few terms to get a feeling for what happens to a n as
n gets large:
8
a 1 − 1 a 2 − 1 ? 2
2 ? 2
a n − 1 ? 2 ? 3 ? ∙ ∙ ∙ ? n
n ? n ? n ? ∙ ∙ ∙ ? n
a 3 − 1 ? 2 ? 3
3 ? 3 ? 3
It appears from these expressions and the graph in Figure 10 that the terms are decreasing
and perhaps approach 0. To confirm this, observe from Equation 8 that
a n − S 1 2 ? 3 ? ∙ ∙ ∙ ? n
n n ? n ? ∙ ∙ ∙ ? nD
n
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