Solução_Calculo_Stewart_6e

F.
TX.10SECTION 11.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS ¤ 469
29. Clearly the series cannot converge if p ≥− 1 2
, because then lim n(1 +
n→∞ n2 ) p 6=0. So assume p1000.
35. f(x) =1/(2x +1) 6 is continuous, positive, and decreasing on [1, ∞), so the Integral Test applies. Using (2),
37.
R n ≤
∞
n
(2x +1) −6 dx = lim
t→∞
t
−1
1
=
. To be correct to five decimal places, we want
10(2x +1) 5 n
10(2n +1)
5
1
10(2n +1) ≤ 5
√ ⇔ (2n +1) 5 ≥ 20,000 ⇔ n ≥ 1 5
5 10 6 2 20,000 − 1 ≈ 3.12,sousen =4.
s 4 =
4
n=1
1
(2n +1) = 1 6 3 + 1 6 5 + 1 6 7 + 1 ≈ 0.001 446 ≈ 0.00145.
6 96 ∞
n −1.001 = ∞
n=1
R n ≤
∞
n
n=1
1
is a convergent p-series with p =1.001 > 1. Using(2),weget
n1.001 x
x −1.001 −0.001 t
dx = lim
= −1000 lim
t→∞ −0.001
n
t→∞
1
x 0.001 t
n
= −1000 − 1
= 1000
n 0.001 n . 0.001
We want R n < 0.000 000 005 ⇔ 1000
n 0.001 < 5 × 10−9 ⇔ n 0.001 > 1000
5 × 10 −9 ⇔
n> 2 × 10 11 1000 =2 1000 × 10 11,000 ≈ 1.07 × 10 301 × 10 11,000 =1.07 × 10 11,301 .