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

58 0 CHAPTER 11 INFINITE SEQUENCES AND SERIES
83. Suppose on the contrary that l:(an + bn) converges. Then l:(an + bn) and I: an are convergent series. So by
Theorem 8(iii), I: [(a,. + bn) - a,.] would also b~ convergent. But I: [(an + bn) -an] = I: bn, a contradiction, since
I: bn is given to be divergent.
85. The partial sums {sn} form an increasing sequence, since s,. - Sn- 1 =an > 0 for all n. Also, the sequence {sn} is bounded
since Sn ::::; 1000 for all n. So by the Monotonic Sequence Theorem, the sequence of partial sums converges, that is, the series
I: an is convergent.
87. (a) At the first step, only the interval ( ~ , ~) (length ~) is removed. At the second step, we remove the intervals ( i, ~) and
( ~ ' ~).which have a total length of2 · Ut At the third step 1 we remove 2 2 intervals, each of length (~ ) 3 • In general,
at the nth step we remove 2n- 1 intervals, each oflength ( ~) n, for a length of 2"- 1 · ( ~)"' = ~ ( ~) n- 1 • Thus, the total
length of all removed intervals is f ~ ( t) n- 1 = 1 :./; 13
n=l
the n th step, the leftmost interval that is removed is ( ( 1
= 1 [geometric se~ies with a = t and 7' = t] . Notice. that at
)" , ( j )"),so we never remov~ 0, and 0 is in the Cantor set. Also,
the rightmost interval removed is (1 -
12127 d B
are 3• 3•·9• 9• 9• an 9·
(j)", 1 - (tf'), so 1 is never removed. Some other numbers in the Cantor set
(b) The area removed at the first step is ~; at the second step, 8 · ( ~) 2 ; at tbe third step, {8) 2 · ( i) 3 • In general, the area
removed at t.he nth step is {8)"-1
00
1
(8)"- 1 1 / 9
n>;1 9 9 = 1 - 8/ 9 = 1.
a r = H ~r-l. so the totat area of all removed squares is
oo n 1 1 1 2 5 5 3 23
89. (a) For n>;1 (n + 1)!' 81 = 1. 2 = 2' S2 = 2 + ~ = 6' sa = 6 + 1 . 2. 3. 4 24,
23 4 119 . (n + 1}!-1
1 ) 1
s 4 = - + = -.The denommators are (n + 1}!, so a guess would be sn = (
24 1 · 2 · 3 · 4 · 5 120 n + .
1 21 - 1 (k + 1)! - 1
(b) For n = 1, 8 1 =
2 = ~·so the formula holds for n = 1. Assume 8k = (k + )! . Then
1
(k + 1}! - 1 k + 1 (k + 1}! - 1 k + 1 (k + 2)! - (k + 2) + k + 1
Sk+l = (k + 1)! + (k + 2)! = (k + 1)! + (k + l)!(k + 2) = (k + 2)!
- (k + 2)! -1
- (It+ 2)!
Thus, the formula is true for n = k + 1. So by induction, the guess is correct.
. · . (n + 1)! - 1 . [ 1 ] oo n
(c) lim 8n = lim ( )I = hm 1 - ( )' = 1 and so L: (
n-oo ,._00 n + 1 . n-oo n + 1 . "=1 n + 1
)! = 1.
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