Solução_Calculo_Stewart_6e

F.
464 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES
TX.10
49.
51.
∞
4 n x n
= ∞ (4x) n is a geometric series with r =4x, so the series converges ⇔ |r| < 1 ⇔ 4 |x| < 1 ⇔
n=0
n=0
|x| < 1 4 . In that case, the sum of the series is 1
1 − 4x .
∞
n=0
cos n x
is a geometric series with first term 1 and ratio r = cos x
|cos x|
,soitconverges ⇔ |r| < 1. But|r| =
2 n 2 2
for all x. Thus, the series converges for all real values of x and the sum of the series is
1
1 − (cos x)/2 = 2
2 − cos x .
53. After defining f, Weuseconvert(f,parfrac); in Maple, Apart in Mathematica, or Expand Rational and
Simplify in Derive to find that the general term is 3n2 +3n +1
= 1 (n 2 + n) 3 n − 1
.Sothenth partial sum is
3 (n +1)
3
s n =
n 1
k − 1
=
1 − 1 1
+
3 (k +1) 3 2 3 2 − 1
1
+ ···+
3 3 3 n − 1
1
=1−
3 (n +1) 3 (n +1) 3
k=1
The series converges to lim
n→∞ s n =1. This can be confirmed by directly computing the sum using sum(f,1..infinity);
(in Maple), Sum[f,{n,1,Infinity}] (in Mathematica), or Calculus Sum (from 1 to ∞)andSimplify (in Derive).
55. For n =1, a 1 =0since s 1 =0.Forn>1,
Also,
∞
n=1
a n = s n − s n−1 = n − 1 (n − 1) − 1 (n − 1)n − (n +1)(n − 2) 2
− = =
n +1 (n − 1) + 1 (n +1)n
n(n +1)
a n = lim s 1 − 1/n
n = lim
n→∞ n→∞ 1+1/n =1.
57. (a) The first step in the chain occurs when the local government spends D dollars. The people who receive it spend a fraction c
of those D dollars, that is, Dc dollars. Those who receive the Dc dollars spend a fraction c of it, that is, Dc 2 dollars.
Continuing in this way, we see that the total spending after n transactions is
≤ 1 2
S n = D + Dc + Dc 2 + ···+ Dc n–1 = D(1 − cn )
1 − c
by (3).
(b) lim S D(1 − c n )
n = lim
= D
n→∞ n→∞ 1 − c 1 − c lim (1 − n→∞ cn )=
D
1 − c
since 0 1 or 1+c0 or c