Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

52 0 CHAPTER 11 INFINITE SEQUENCES AND SERIES
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3. I: an = lim s, = lim [2- 3(0.8)n) = lim 2 - 3 lim (0.8)" = 2 - 3(0) = 2
n=l n-oo n-.oo n--too n-+oo
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1 1 1 1
5. For I: 3 • a,,= 3· 81 = a1 = 13 = 1, 82 = s1 + a2 = 1 + 23
= 1.125, sa = 82 + aa ~ 1.1620,
n=l n n
s~ = 83 +a~ ~ 1.1777, ss = 84 +as~ 1.1857, so = ss + ao ~ 1.1903, s 7 = 8G + a 7 ~ 1.1932, and
ss = sr + as ~ 1.1952. It appears that the series is convergent.
'
oo n n 1 2
7. For I: ~ ·a,, = ----r.:;· St = a1 i;= - -- = 0.5 82 = St + a2 = 0.5 + ~ ~ 1.3284,
n=l 1 + yn 1 + yn 1 + J1 ' 1 + v2
s3 = s2 +as~ 2.4265, S4 = sa+ a4 :::::: 3.7598, 8::, = 84 + a 5 :::::: 5.3049, 8 6 = s 5 + a 6 :::::: 7.0443,
sr = so+ ar:::::: 8.9644, ss = 87 +as~ 11.0540. It appears that the series is divergent.
9.
n
8n
1 - 2.40000
2 - 1.92000
3 - 2.01600
4 - 1.99680
5 - 2.00064
Ot----+--;---+---+----..----+--i ll
6 - 1.99987 - 3
7 -2.00003
From the graph and the table, it seems that the series converges to - 2. In fact, it is a geometric
8 -1.99999
· ·th
9 - 2.00000
2 d I · · . ~ 12 -2.4 -2.4
sen es w1 a= - .4 an r =-&·so tis sum IS 6 ( -
5 )n = ( 1) = -- = - 2.
. n=l 1 - - & 1.2
10 -2.00000
Note that the dot corresponding to n = 1 is part of both {an} and { Sn}.
TI-86 Note: To graph {an} and { sn}, set your calculator to Param mode and Draw Dot mode. (Draw Dot is under
GRAPH, MORE, FORMT (F3).) Now under E (t) = make the assignments: xt1=t, yt1=12/ ( -5) -t, xt2=t,
yt2=surn seq ( ytl, t, 1, t , 1) . (sum and seq are under LIST, OPS (F5), MORE.) Under WIND use
1 , 10, 1, 0, 1 0, 1, - 3, 1, 1 toobtain a graph s im ilar to the one above. Then use TRACE (F4) to see the values.
11.
n
Sn.
10/------------~
1
2
3
4
5
6
7
8
9
10
0.44721
1.15432
1.98637
2.88080
3.80927
4.75796
5.71948
6.68962
7.66581
8.64639
0....___....__ _
__..__..__
__..___..__, 11
The series f: ~ diverges, since its terms do not approach 0.
n=l n 2 + 4
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