Solução_Calculo_Stewart_6e

F.
TX.10SECTION 11.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS ¤ 467
1
5. The function f(x) = is continuous, positive, and decreasing on [1, ∞), so the Integral Test applies.
(2x +1)
3
∞
1
t
1
dx = lim
dx = lim − 1 t
1
1
= lim −
1 (2x +1)
3 t→∞
1 (2x +1)
3 t→∞ 4 (2x +1) 2 t→∞
1
4(2t +1) + 1
= 1
2 36 36 .
Since this improper integral is convergent, the series ∞
n=1
1
is also convergent by the Integral Test.
(2n +1)
3
7. f(x) =xe −x is continuous and positive on [1, ∞). f 0 (x) =−xe −x + e −x = e −x (1 − x) < 0 for x>1,sof is decreasing
on [1, ∞). Thus, the Integral Test applies.
∞
1
xe −x dx = lim
b→∞
b
1 xe−x dx = lim
b→∞
−xe −x − e −x b
1
[by parts] = lim
b→∞
[−be −b − e −b + e −1 + e −1 ]=2/e
since lim
b→∞
be −b = lim
b→∞
(b/e b ) H = lim
b→∞
(1/e b )=0and lim
b→∞
e −b =0. Thus, ∞
n=1 ne−n converges.
9. The series ∞
n=1
for if it converged, then ∞
1
n is a p-series with p =0.85 ≤ 1,soitdivergesby(1).Therefore,theseries ∞
0.85
n=1
11. 1+ 1 8 + 1 27 + 1 64 + 1
125 + ···= ∞
13. 1+ 1 3 + 1 5 + 1 7 + 1 9 + ··· = ∞
15.
1
would have to converge [by Theorem 8(i) in Section 11.2].
n0.85 n=1
n=1
n=1
1
.Thisisap-series with p =3> 1, so it converges by (1).
n3 1
2n − 1 . The function f(x) = 1
2x − 1 is
continuous, positive, and decreasing on [1, ∞), so the Integral Test applies.
∞
1
diverges.
∞
n=1
1
dx = lim
2x − 1 t→∞
5 − 2 √ n
n 3
=5 ∞
n=1
t
1
2
must also diverge,
n0.85 1
dx = lim 1
2x − 1 ln |2x − 1| t
= 1 lim (ln(2t − 1) − 0) = ∞,sotheseries ∞ 1
t→∞ 2 1 2 t→∞ n=1 2n − 1
1
n − 2 ∞
3 n=1
with p =3> 1 and p =
5
> 1 ∞
2
. Thus,
1
n by Theorem 11.2.8, since ∞
5/2
n=1
5 − 2 √ n
n 3
converges.
n=1
1
n and ∞
3 n=1
1
both converge by (1)
n5/2 17. The function f(x) = 1 is continuous, positive, and decreasing on [1, ∞),sowecanapplytheIntegralTest.
x 2 +4
∞
1
t
t
1
1 x
dx = lim
dx = lim
1 x 2 +4 t→∞
1 x 2 +4 t→∞ 2 tan−1 = 1
t 1
2
1
2 lim tan −1 − tan −1
t→∞ 2
2
= 1 π 1
2 2 2
− tan−1
19.
Therefore, the series ∞
∞
n=1
ln n
n 3
= ∞
n=2
n=1
1
n 2 +4 converges.
ln n ln 1
ln x
since =0. The function f(x) = is continuous and positive on [2, ∞).
n3 1 x 3
f 0 (x) = x3 (1/x) − (ln x)(3x 2 )
= x2 − 3x 2 ln x
= 1 − 3lnx < 0 ⇔ 1 − 3lnx 1
(x 3 ) 2 x 6
x 4 3
⇔