Solução_Calculo_Stewart_6e

F.
468 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES
TX.10
x>e 1/3 ≈ 1.4,sof is decreasing on [2, ∞), and the Integral Test applies.
∞
2
ln x
x 3
converges.
t
ln x
dx = lim
t→∞
2 x 3
()
dx = lim − ln x
t→∞ 2x − 1 t
= lim − 1
2 4x 2 t→∞
1
4t (2 ln t +1)+ 1
()
= 1 2 4 4 ,sotheseries ∞ ln n
n=2 n 3
(): u =lnx, dv = x −3 dx ⇒ du =(1/x) dx, v = − 1 2 x−2 ,so
ln x
dx = − 1 2 x−2 ln x − − 1 2 x−2 (1/x) dx = − 1 2 x−2 ln x + 1 2
x 3
(): lim −
t→∞
2lnt +1 H 2/t
= − lim
4t 2
t→∞ 8t = − 1 4
lim 1
t→∞ t =0. 2
x −3 dx = − 1 2 x−2 ln x − 1 4 x−2 + C.
21. f(x) = 1
x ln x is continuous and positive on [2, ∞), and also decreasing since f 0 (x) =− 1+lnx < 0 for x>2, so we can
x 2 (ln x)
2
∞
1
use the Integral Test.
dx = lim
2 x ln x [ln(ln t→∞ x)]t 2 = lim [ln(ln t) − ln(ln 2)] = ∞,sotheseries ∞ 1
t→∞ n=2 n ln n diverges.
23. The function f(x) =e 1/x /x 2 is continuous, positive, and decreasing on [1, ∞), so the Integral Test applies.
[g(x) =e 1/x is decreasing and dividing by x 2 doesn’t change that fact.]
∞
t
e 1/x
f(x) dx = lim dx = lim
−e 1/x t
= − lim (e 1/t
− e) =−(1 − e) =e − 1, sotheseries ∞ e 1/n
1
t→∞
1 x 2 t→∞
1 t→∞ n=1 n 2
converges.
25. The function f(x) = 1 is continuous, positive, and decreasing on [1, ∞), so the Integral Test applies. We use partial
x 3 + x
fractions to evaluate the integral:
∞
1
so the series ∞
n=1
1
dx = lim
x 3 + x t→∞
t
1
= lim ln
t→∞
1
n 3 + n converges.
1
x −
x
dx = lim ln x − 1 t
1+x 2 t→∞ 2 ln(1 + x2 ) = lim ln
t→∞
1
t
√ − ln 1
√ = lim ln
1+t
2 2 t→∞
27. We have already shown (in Exercise 21) that when p =1the series ∞
n=2
1
1+1/t
2 + 1 2 ln 2
= 1 2 ln 2
t
x
√
1+x
2
1
1
diverges, so assume that p 6= 1.
n(ln n)
p
1
f(x) =
x(ln x) is continuous and positive on [2, ∞),andf 0 (x) =−
p +lnx
p x 2 (ln x) < 0 if p+1 x>e−p ,sothatf is eventually
decreasing and we can use the Integral Test.
∞
2
1
(ln x)
1−p
dx = lim
x(ln x)
p t→∞ 1 − p
t
2
(ln t)
1−p
[for p 6= 1] = lim
−
t→∞ 1 − p
This limit exists whenever 1 − p1, so the series converges for p>1.
(ln 2)1−p
1 − p