Solução_Calculo_Stewart_6e

F.
SECTION TX.10 11.6 ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS ¤ 475
29. b 7 = 72
=0.000 004 9, so
107 31.
∞
n=1
(−1) n−1 n 2
10 n ≈ s 6 = 6
n=1
(−1) n−1 n 2
= 1
10 − 4 + 9 − 16 + 25 − 36
n 10 100 1000 10,000 100,000 1,000,000
=0.067 614. Addingb7 to s6
does not change the fourth decimal place of s 6 , so the sum of the series, correct to four decimal places, is 0.0676.
∞
n=1
(−1) n−1
n
underestimate, since ∞
=1− 1 2 + 1 3 − 1 4 + ···+ 1
49 − 1
50 + 1
51 − 1 + ···. The 50th partial sum of this series is an
52
n=1
(−1) n−1
The result can be seen geometrically in Figure 1.
n
1
= s 50 +
51 52
− 1 1
+
53 − 1
+ ···, and the terms in parentheses are all positive.
54
33. Clearly b n = 1 is decreasing and eventually positive and lim
n + p b n =0for any p. So the series converges (by the
n→∞
Alternating Series Test) for any p for which every b n is defined, that is, n + p 6= 0for n ≥ 1,orp is not a negative integer.
35.
b2n = 1/(2n) 2 clearlyconverges(bycomparisonwiththep-series for p =2). So suppose that (−1) n−1 b n
converges. Then by Theorem 11.2.8(ii), so does (−1) n−1 b n + b n
=2
1+
1
3 + 1 5 + ··· =2 1
2n − 1 .Butthis
diverges by comparison with the harmonic series, a contradiction. Therefore, (−1) n−1 b n must diverge. The Alternating
Series Test does not apply since {b n} is not decreasing.
11.6 Absolute Convergence and the Ratio and Root Tests
1. (a) Since lim
(b) Since lim
n→∞ an+1
a n
n→∞ an+1
a n
therefore convergent).
(c) Since lim
n=0
n→∞ an+1
a n
=8> 1, part (b) of the Ratio Test tells us that the series a n is divergent.
=0.8 < 1, part (a) of the Ratio Test tells us that the series a n is absolutely convergent (and
=1, the Ratio Test fails and the series a n might converge or it might diverge.
∞ (−10) n
3.
. Using the Ratio Test, lim
n!
n→∞ a n+1 = lim
n→∞ (−10)n+1 n! · = lim
(n +1)! (−10) n n→∞ −10
n +1 =0< 1,sotheseriesis
5.
absolutely convergent.
∞
n=1
(−1) n+1
4√ n
is conditionally convergent.
a n
converges by the Alternating Series Test, but
∞
n=1
1
4√ n
is a divergent p-series p = 1 4 ≤ 1 ,sothegivenseries