Solução_Calculo_Stewart_6e

F.
466 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES
TX.10
73. (a) At the first step, only the interval 1
, 2
3 3 (length
1
) is removed. At the second step, we remove the intervals 1
, 2
3 9 9 and
7 , 8
9 9 , which have a total length of 2 · 1
2
. At the third step, we remove 2 2 intervals, each of length 1
3
3
3
. In general,
at the nth step we remove 2 n−1 intervals, each of length
1 n
,foralengthof2 n−1 ·
1 n
= 1 2
n−1
. Thus, the total
3
3 3 3
length of all removed intervals is ∞
n=1
the nth step, the leftmost interval that is removed is
1 n
, 2
3
the rightmost interval removed is 1 − 2
3
are 1 3 , 2 3 , 1 9 , 2 9 , 7 9 ,and 8 9 .
1 2
n−1
= 1/3 =1 geometric series with a = 1 and r = 2
3 3
1 − 2/3 3 3 . Notice that at
n
3 ,soweneverremove0,and0 is in the Cantor set. Also,
n
, 1 −
1 n
3 ,so1 is never removed. Some other numbers in the Cantor set
(b) The area removed at the first step is 1 ;atthesecondstep,8 ·
1 2
;atthethirdstep,(8) 2 ·
1 3
. In general, the area
9 9
9
removed at the nth step is (8) n−1
1 n
9
= 1 8
n−1
9 9
, so the total area of all removed squares is
∞ n−1
1 8
= 1/9
9 9 1 − 8/9 =1.
n=1
75. (a) For ∞
n=1
n
(n +1)! , s1 = 1
1 · 2 = 1 2 , s2 = 1 2 + 2
1 · 2 · 3 = 5 6 , s3 = 5 6 + 3
1 · 2 · 3 · 4 = 23
24 ,
s 4 = 23
24 + 4
1 · 2 · 3 · 4 · 5 = 119
120 . The denominators are (n +1)!,soaguesswouldbes n =
(b) For n =1, s 1 = 1 2 = 2! − 1 , so the formula holds for n =1. Assume s k =
2!
s k+1 =
=
(k +1)!− 1
(k +1)!
(k +2)!− 1
(k +2)!
(k +1)!− 1
.Then
(k +1)!
(n +1)!− 1
.
(n +1)!
+ k +1 (k +1)!− 1 k +1 (k +2)!− (k +2)+k +1
= +
=
(k +2)! (k +1)! (k +1)!(k +2) (k +2)!
Thus, the formula is true for n = k +1. So by induction, the guess is correct.
(n +1)!− 1
(c) lim sn = lim
= lim 1 −
n→∞ n→∞ (n +1)! n→∞
1
=1and so ∞
(n +1)!
n=1
n
(n +1)! =1.
11.3 The Integral Test and Estimates of Sums
1. The picture shows that a 2 = 1 2
2 < 1
dx,
1.3 x1.3 a 3 = 1
3 1.3 < 3
2
1
x dx, and so on, so ∞
1.3
1
n=2
1 ∞
n < 1.3
1
1
dx. The
x1.3 integral converges by (7.8.2) with p =1.3 > 1, so the series converges.
3. The function f(x) =1/ 5√ x = x −1/5 is continuous, positive, and decreasing on [1, ∞), so the Integral Test applies.
∞
t
x −1/5 t
dx = lim
1
t→∞ 1 x−1/5 5
dx = lim
t→∞ 4 x4/5 5
= lim
1 t→∞ 4 t4/5 − 5 = ∞,so ∞ 1/ 5√ n diverges.
4
n=1