Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 11.4 THE COMPARISON TESTS ¤ 471
17. Use the Limit Comparison Test with a n =
a n
lim = lim
n→∞ b n n→∞
∞
n=1
1
√
n2 +1 .
n
√
n2 +1 = lim
n→∞
1
√
n2 +1 and bn = 1 n :
1
1+(1/n2 ) =1> 0. Since the harmonic series ∞
19. Use the Limit Comparison Test with a n = 1+4n
1+3 n and b n = 4n
3 n :
a n
lim = lim
n→∞ b n n→∞
1+4 n
1+3 n 1+4 n
4 n = lim
n→∞ 1+3 · 3n
n 4 = lim 1+4 n
·
n n→∞ 4 n
3 n
Since the geometric series b n =
4 n
diverges, so does ∞ 3
n=1
1+4 n
1+3 > 1+4n
n 3 n +3 > 4n
n 2(3 n ) = 1 n 4
or use the Test for Divergence.
2 3
21. Use the Limit Comparison Test with a n =
√ n +2
1
and
2n 2 bn =
+ n +1 n : 3/2
a n n √ 3/2 n +2
lim = lim
n→∞ b n n→∞ 2n 2 + n +1 = lim (n √ 3/2 n +2)/(n √ 3/2 n )
n→∞
Since ∞
n=1
(2n 2 + n +1)/n 2 = lim
n→∞
1
n is a convergent p-series p = 3 > 1
,theseries ∞ 3/2 2
n=1
3 n
1
1+3 = lim
n n→∞ 4 +1 ·
n
1
diverges, so does
n
1
=1> 0
1
3 +1 n
1+4 n
. Alternatively, use the Comparison Test with
1+3n n=1
23. Use the Limit Comparison Test with a n = 5+2n
(1 + n 2 ) 2 and bn = 1 n 3 :
a n n 3 (5 + 2n)
lim = lim
n→∞ b n n→∞ (1 + n 2 ) = lim 5n 3 +2n 4
2 n→∞ (1 + n 2 ) · 1/n 4
2 1/(n 2 ) = lim
2 n→∞
p-series [p =3> 1], the series ∞
25. If a n =
so ∞
n=1
n=1
5+2n
also converges.
(1 + n 2 )
2
1+n + n2
√
1+n2 + n and 6 bn = 1 a n
,then lim = lim
n n→∞ b n n→∞
1+2/n
2+1/n +1/n 2 = √
1
2 = 1 2 > 0.
√ n +2
also converges.
2n 2 + n +1
5
+2 n
1
+1 2 =2> 0. Since ∞
n 2 n=1
n + n 2 + n 3
√
1+n2 + n 6 = lim
n→∞
1+n + n 2
√
1+n2 + n diverges by the Limit Comparison Test with the divergent harmonic series ∞
6
27. Use the Limit Comparison Test with a n =
∞
e −n
= ∞
n=1
n=1
1+ 1 2
e −n and b n = e −n a n
: lim
n n→∞
1
e is a convergent geometric series |r| = 1 < 1
,theseries ∞ n e
29. Clearly n! =n(n − 1)(n − 2) ···(3)(2) ≥ 2 · 2 · 2 ·····2 · 2=2 n−1 ,so 1 n! ≤ 1
2 n−1 . ∞
series |r| = 1 2 < 1 ,so ∞
n=1
1
converges by the Comparison Test.
n!
n=1
b n
1
is a convergent
n3 1/n 2 +1/n +1
=1> 0,
1/n6 +1/n 4 +1
n=1
1
n .
= lim 1+ 1 2
=1> 0. Since
n→∞ n
1+ 1 n
2
e −n also converges.
n=1
1
is a convergent geometric
2n−1