Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 11.7 STRATEGY FOR TESTING SERIES ¤ 479
9.
∞
k 2 e −k
= ∞
k=1
a k
k=1
lim
k→∞
a k+1 = lim
(k +1)2
k→∞
e k+1
k 2
. Using the Ratio Test, we get
ek · ek
k 2
= lim
k→∞
k +1
k
2
· 1 =1 2 · 1
e e = 1 < 1, so the series converges.
e
11. b n = 1
n ln n > 0 for n ≥ 2, {b n} is decreasing, and lim b
n =0,sothegivenseries ∞ (−1) n+1
converges by the
n→∞ n=2 n ln n
Alternating Series Test.
13. lim
n→∞ a n+1 = lim
n→∞ 3n+1 (n +1) 2
·
(n +1)!
a n
convergesbytheRatioTest.
a n
n!
3 n n 2
= lim
n→∞
15. lim
n→∞ a n+1 = lim
(n +1)!
n→∞ 2 · 5 · 8 ·····(3n +2)[3(n +1)+2]
so the series ∞
n=0
n!
2 · 5 · 8 ·····(3n +2) convergesbytheRatioTest.
3(n +1) 2 n +1
=3 lim =0< 1,sotheseries ∞ 3 n n 2
(n +1)n2 n→∞ n 2
n=1 n!
· 2 · 5 · 8 ·····(3n +2)
n!
n=1
= lim
n→∞
n +1
3n +5 = 1 3 < 1,
17. lim
n→∞ 21/n =2 0 =1,so lim
n→∞ (−1)n 2 1/n
does not exist and the series ∞ (−1) n 2 1/n diverges by the Test for Divergence.
19. Let f(x) = ln √ x .Thenf 0 (x) = 2 − ln x < 0 when ln x>2 or x>e 2 ,so ln √ n is decreasing for n>e 2 .
x 2x 3/2
n
21.
ln n
By l’Hospital’s Rule, lim √ = lim
n→∞ n
n→∞
Alternating Series Test.
∞
n=1
(−2) 2n
n n
1/n
1/ 2 √ 2
= lim √ =0,sotheseries ∞ (−1) n ln n
√ converges by the
n→∞
n
n n=1 n
=
n=1 ∞ n 4
n
4
. lim |an | = lim =0< 1, so the given series is absolutely convergent by the Root Test.
n n→∞
n→∞ n
1
23. Using the Limit Comparison Test with a n =tan and b n =
n
1 n ,wehave
a n tan(1/n) tan(1/x) H sec 2 (1/x) · (−1/x 2 )
lim = lim = lim = lim
= lim
n→∞ b n n→∞ 1/n x→∞ 1/x x→∞ −1/x 2
x→∞ sec2 (1/x) =1 2 =1> 0. Since
∞
∞
b n is the divergent harmonic series, a n is also divergent.
n=1
25. Use the Ratio Test. lim
27.
converges.
n→∞ an+1
a n
= lim
n→∞
n=1
(n +1)!
e (n+1)2
∞
ln x
dx = lim − ln x
2 x2 t→∞ x − 1 t
x
1
k ln k
(k +1) < k ln k = ln k
3 k 3 k ,thegivenseries ∞
2 k=1
· en2
n!
= lim
n→∞
(n +1)n! · e n2
e n2 +2n+1
n!
[using integration by parts] H =1.So ∞
n=1
ln n
n 2
k ln k
3
converges by the Comparison Test.
(k +1)
n +1
= lim
n→∞ e =0< 1,so ∞ n!
2n+1 n=1 e n2
converges by the Integral Test, and since