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

706 Chapter 11 Infinite Sequences and Series
83. (a) Fibonacci posed the following problem: Suppose that
rabbits live forever and that every month each pair
produces a new pair which becomes productive at
age 2 months. If we start with one newborn pair, how
many pairs of rabbits will we have in the nth month?
Show that the answer is f n, where h f nj is the Fibonacci
sequence defined in Example 3(c).
(b) Let a n − f n11yf n and show that a n21 − 1 1 1ya n22 .
Assuming that ha nj is convergent, find its limit.
84. (a) Let a 1 − a, a 2 − f sad, a 3 − f sa 2d − f s f sadd, . . . ,
a n11 − f sa nd, where f is a continuous function. If
lim n l ` a n − L, show that f sLd − L .
(b) Illustrate part (a) by taking f sxd − cos x, a − 1, and
estimating the value of L to five decimal places.
; 85. (a) Use a graph to guess the value of the limit
n 5
lim
n l ` n!
(b) Use a graph of the sequence in part (a) to find the
smallest values of N that correspond to « − 0.1 and
« − 0.001 in Definition 2.
86. Use Definition 2 directly to prove that lim n l ` r n − 0
when | r | , 1.
87. Prove Theorem 6.
[Hint: Use either Definition 2 or the Squeeze Theorem.]
88. Prove Theorem 7.
89. Prove that if lim n l ` a n − 0 and hb nj is bounded, then
lim n l ` sa nb nd − 0.
90. Let a n −S1 1 1 nDn.
(a) Show that if 0 < a , b, then
b n11 2 a n11
, sn 1 1db n
b 2 a
(b) Deduce that b n fsn 1 1da 2 nbg , a n11 .
(c) Use a − 1 1 1ysn 1 1d and b − 1 1 1yn in part (b) to
show that ha nj is increasing.
(d) Use a − 1 and b − 1 1 1ys2nd in part (b) to show
that a 2n , 4.
(e) Use parts (c) and (d) to show that a n , 4 for all n.
(f) Use Theorem 12 to show that lim n l ` s1 1 1ynd n exists.
(The limit is e. See Equation 3.6.6.)
91. Let a and b be positive numbers with a . b. Let a 1 be their
arithmetic mean and b 1 their geometric mean:
a 1 − a 1 b
2
Repeat this process so that, in general,
a n11 −
an 1 bn
2
b 1 − sab
b n11 − sa nb n
(a) Use mathematical induction to show that
a n . a n11 . b n11 . b n
(b) Deduce that both ha nj and hb nj are convergent.
(c) Show that lim n l ` a n − lim n l ` b n. Gauss called the
common value of these limits the arithmetic-geometric
mean of the numbers a and b.
92. (a) Show that if lim n l ` a 2n − L and lim n l ` a 2n11 − L,
then ha nj is convergent and lim n l ` a n − L.
(b) If a 1 − 1 and
a n11 − 1 1 1
1 1 a n
find the first eight terms of the sequence ha nj. Then use
part (a) to show that lim n l ` a n − s2 . This gives the
continued fraction expansion
s2 − 1 1
2 1
1
1
2 1 ∙ ∙ ∙
93. The size of an undisturbed fish population has been modeled by
the formula
p n11 −
bpn
a 1 p n
where p n is the fish population after n years and a and b are
positive constants that depend on the species and its environment.
Suppose that the population in year 0 is p 0 . 0.
(a) Show that if h p nj is convergent, then the only possible
values for its limit are 0 and b 2 a.
(b) Show that p n11 , sbyadp n.
(c) Use part (b) to show that if a . b, then lim n l ` p n − 0;
in other words, the population dies out.
(d) Now assume that a , b. Show that if p 0 , b 2 a, then
h p nj is increasing and 0 , p n , b 2 a. Show also that
if p 0 . b 2 a, then h p nj is decreasing and p n . b 2 a.
Deduce that if a , b, then lim n l ` p n − b 2 a.
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