7250-Resolução(Alguns)2

Prévia do material em texto

Resoluc¸a˜o
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.13. (a) Multiplicando sn por
1
5


sn = 2 +
2
5
+
2
52
+ ...+
2
5n−1
sn
5
=
2
5
+
2
52
+
2
53
+ ...+
2
5n
Vamos calcular o termo sn. Subtraindo sn − sn
5
, temos
sn − sn
5
= 2− 2
5n
sn
(
1− 1
5
)
= 2− 2
5n
sn
(
4
5
)
= 2− 2
5n
sn =
10
4
− 10
4.5n
=
5
2
− 5
2.5n
=
5n+1 − 5
2.5n
(b) Ana´logo ao anterior.
(c)
uk =
1
(k + 1)(k + 2)
=
A
(k + 1)
+
B
(k + 2)
=
A(k + 2) +B(k + 1)
(k + 1)(k + 2)
=
(A+B)k + 2A+B
(k + 1)(k + 2)
Logo A+B = 0 e 2A+B = 1 que resulta em A = 1 e B = −1. E temos:
sn =
n∑
k=1
1
(k + 1)(k + 2)
=
n∑
k=1
1
(k + 1)
− 1
(k + 2)
=
(
1
2
− 1
3
)
+
(
1
3
− 1
4
)
+
(
1
4
− 1
5
)
+ ...+
(
1
n− 1 −
1
n
)
=
1
2
(
−1
3
+
1
3
)
−
(
1
4
+
1
4
)
−
(
1
5
+ ...+
1
n− 1
)
− 1
n
=
1
2
− 1
n
(d) Ana´logo ao (a)
(e) Ana´logo ao (c)
14. 15.
16.
8
17. (a) Note que:
x− x3 + x5 − x7 + x9 − ... = 1.x1 + (−1).x3 + (1).x5 + (−1).x7 + (1).x9 − ...
= (−1)0.x2.0+1 + (−1)1.x2.1+1 + (−1)2.x2.2+1 + (−1)3.x2.3+1 + (−1)4.x2.4+1 + ...
=
+∞∑
k=0
(−1)kx2k+1
=
+∞∑
k=0
(−1)kx2kx1
=
+∞∑
k=0
(−1)k(x2)kx1
=
+∞∑
k=0
x1(−x2)k
E temos uma se´rie geome´trica com a = x e r = −x2. Para que a se´rie convirja devemos ter:
| − x2| < 1 ⇒ x2 < 1
⇒ x2 − 1 < 0
⇒ (x− 1)(x+ 1) < 0
⇒


(x− 1) > 0 e (x+ 1) < 0
ou
(x− 1) > 0 e (x+ 1) < 0
⇒


x > 1 e x < −1 que e´ imposs´ıvel
ou
x > 1 e x < −1
Logo a se´rie converge se −1 < x < 1 e diverge se −1 > x ou x > 1.
18. 19. 20. 21. 22. 23.
24.
25. (a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
+∞∑
k=1
(−1)k 2
k
k3
Crite´rio da raiz.
lim
k→+∞
k
√∣∣∣∣(−1)k 2kk3
∣∣∣∣ = limk→+∞ k
√
|(−1)k| |2
k|
|k3| = limk→+∞
k
√
1
2k
k3
= lim
k→+∞
k
√
2k
k3
= lim
k→+∞
k
√
2k
k
√
k3
= lim
k→+∞
2
(k3)
1
k
= lim
k→+∞
2
k
3
k
=
2
lim
k→+∞
k
3
k
Considere a func¸a˜o f(x) = x
3
x para x > 0, aplicando ln temos:
ln f(x) = lnx
3
x =
3
x
lnx
9
Aplicando e temos:
eln f(x) = e
3
x
lnx
f(x) = e
3
x
lnx
Aplicando o limite temos:
lim
x→+∞
f(x) = lim
x→+∞
e
3
x
lnx = e
lim
x→+∞
3
x
lnx
= e
lim
x→+∞
3 ln x
x
Aplicando L’Hopital em lim
x→+∞
3 lnx
x
temos:
lim
x→+∞
3 lnx
x
= lim
x→+∞
( 3
x
)
1
=
0
1
= 0
Logo lim
x→+∞
f(x) = e0 = 1, e assim
2
lim
k→+∞
k
3
k
=
2
1
= 2, e lim
k→+∞
k
√∣∣∣∣(−1)k 2kk3
∣∣∣∣ = 2 > 1. Pelo teste
da raiz a se´rie diverge.
.
10