Solução_Calculo_Stewart_6e

F.
SECTION TX.10 11.6 ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS ¤ 477
25. UsetheRatioTestwiththeseries
1 − 1 · 3 + 1 · 3 · 5 − 1 · 3 · 5 · 7 + ···+(−1) n−1 1 · 3 · 5 ·····(2n − 1)
+ ···= ∞ (−1) n−1 1 · 3 · 5 ·····(2n − 1)
.
3! 5! 7!
(2n − 1)!
n=1
(2n − 1)!
lim
n→∞ a n+1 = lim
a n n→∞ (−1)n · 1 · 3 · 5 ·····(2n − 1)[2(n +1)− 1]
(2n − 1)!
·
[2(n +1)− 1]!
(−1) n−1 · 1 · 3 · 5 ·····(2n − 1)
= lim
(−1)(2n +1)(2n − 1)!
n→∞ (2n +1)(2n)(2n − 1)! = lim 1
=0< 1,
n→∞ 2n
27.
so the given series is absolutely convergent and therefore convergent.
∞
n=1
2 · 4 · 6 ·····(2n)
n!
= ∞
n=1
Divergence since lim
n→∞ 2n = ∞.
29. Bytherecursivedefinition, lim
31. (a) lim
n→∞
1/(n +1)3
1/n 3
= lim
n→∞
(b) lim
(n +1)
n→∞ · 2n 2 n+1 n = lim
(c) lim
√ (−3)n · n +1
n→∞
(d) lim
n→∞
33. (a) lim
(2 · 1) · (2 · 2) · (2 · 3) ·····(2 · n)
n!
n→∞ an+1
a n
n→∞
= lim
n→∞
n 3
(n +1) 3 = lim
n→∞
n +1
2n
√
n
= 3 lim
(−3) n−1 n→∞
= ∞
n=1
2 n n!
n!
= ∞ 2 n , which diverges by the Test for
n=1
5n +1
4n +3 = 5 > 1, so the series diverges by the Ratio Test.
4
1 = lim
n→∞ 2 + 1
2n
n
n +1
√
n +1
1+(n +1) · 1+n2 √ = lim
2 n n→∞
n→∞ an+1
a n
series ∞
n=0
1
3
=1. Inconclusive
(1 + 1/n)
1+ 1 n ·
= lim
x n+1
n→∞ (n +1)! · n! = lim
x
x n n→∞ n +1 x n
n!
converges for all x.
= 1 . Conclusive (convergent)
2
1
=3 lim
n→∞ 1+1/n =3.
Conclusive (divergent)
1/n 2 +1
1/n 2 +(1+1/n) 2 =1. Inconclusive
= |x| lim
n→∞
x n
(b) Since the series of part (a) always converges, we must have lim =0by Theorem 11.2.6.
n→∞ n!
35. (a) s 5 = 5
n=1
1
n2 = 1 n 2 + 1 8 + 1
24 + 1 64 + 1
160 = 661 ≈ 0.68854. Nowtheratios
960
r n = a n+1 n2 n
=
a n (n +1)2 = n
form an increasing sequence, since
n+1 2(n +1)
1
= |x|·0=0< 1,sobytheRatioTestthe
n +1
r n+1 − r n = n +1
2(n +2) − n
2(n +1) = (n +1)2 − n(n +2) 1
=
> 0. So by Exercise 34(b), the error
2(n +1)(n +2) 2(n +1)(n +2)
a 6
in using s 5 is R 5 ≤
= 1/ 6 · 2 6
1 − lim
n→∞ rn 1 − 1/2 = 1
192 ≈ 0.00521.
(b) The error in using s n as an approximation to the sum is R n = an+1
1 − 1 2
=
2
(n +1)2 n+1 .WewantR n < 0.00005 ⇔
1
(n +1)2 n < 0.00005 ⇔ (n +1)2n > 20,000. Tofind such an n we can use trial and error or a graph. We calculate
(11 + 1)2 11 =24,576,sos 11 = 11
n=1
1
≈ 0.693109 is within 0.00005 of the actual sum.
n2n