Solução_Calculo_Stewart_6e

F.
478 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES
TX.10
37. (i) Following the hint, we get that |a n| 1 for n ≥ N. Thus,
lim
n→∞ an 6= 0,so ∞
n=1
an diverges by the Test for Divergence.
(iii) Consider ∞
n=1
1
n [diverges] and ∞
n=1
1
n 2
n
[converges]. For each sum, lim |an| =1, so the Root Test is inconclusive.
n→∞
39. (a) Since a n is absolutely convergent, and since a
+
n
≤ |an| and a
−
n
≤ |an| (because a + n and a − n each equal
either a n or 0), we conclude by the Comparison Test that both a + n and a − n must be absolutely convergent.
Or: Use Theorem 11.2.8.
(b) We will show by contradiction that both a + n and a − n must diverge. For suppose that a + n converged. Then so
would a + n − 1 2 an
by Theorem 11.2.8. But
a
+
n − 1 2 an
=
1
2 (an + |an|) − 1 2 an
=
1
2
|an|, which
diverges because a n is only conditionally convergent. Hence, a + n can’t converge. Similarly, neither can a − n .
11.7 Strategy for Testing Series
1.
1
n +3 < 1 n 1 n 3 = for all n ≥ 1.
n 3
converges by the Comparison Test.
∞
n=1
1
3
n
is a convergent geometric series |r| = 1 3 < 1 ,so ∞
n=1
1
n +3 n
n
n
3. lim |an| = lim =1,so lim an = lim
n→∞ n→∞ n +2 n→∞ n→∞ (−1)n n +2 does not exist. Thus, the series ∞ (−1) n n
diverges by
n +2
the Test for Divergence.
5. lim
n→∞ an+1
a n
∞
n=1
n 2 2 n−1
(−5) n
= lim
n→∞
(n +1)2 2 n (−5) n 2(n +1) 2
· = lim
= 2
(−5) n+1 n 2 2 n−1 n→∞ 5n 2 5 lim 1+ 1 2
= 2
n→∞ n 5 (1) = 2 5 < 1,sotheseries
converges by the Ratio Test.
1
7. Let f(x) =
x √ .Thenf is positive, continuous, and decreasing on [2, ∞), so we can apply the Integral Test.
ln x
1 u =lnx,
Since
x √ ln x dx = u −1/2 du =2u 1/2 + C =2 √ ln x + C, wefind
du = dx/x
∞
dx
t
2 x √ ln x = lim dx
t→∞
2 x √ ln x = lim 2 √ t
ln x = lim 2 √ ln t − 2 √
ln 2 = ∞. Since the integral diverges, the
t→∞
2 t→∞
given series ∞ 1
n=2 n √ ln n diverges.
n=1