CalcII 2011 1 res1aP M1

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Universidade Federal de Minas Gerais
Instituto de Cieˆncias Exatas – ICEx
Departamento de Matema´tica
Ca´lculo Diferencial e Integral II
Resoluc¸a˜o da 1a prova - Turma M1 - 13/04/2011
1. Calcule as somas:
(a)
∞∑
n=1
(
5
2n−1
− 1
3n−1
)
, (b)
∞∑
n=1
2n+ 1
n2(n+ 1)2
.
Soluc¸a˜o. (a) 5
∑ 1
2n−1
−
∑ 1
3n−1
=
5
1− 1
2
− 1
1− 1
3
= 2.5− 3
2
=
17
2
.
(b) Como 2n+ 1 = (n+ 1)2 − n2, temos que 2n+ 1
n2(n+ 1)2
=
1
n2
− 1
(n+ 1)2
.
E,
∞∑
n=1
2n+ 1
n2(n+ 1)2
= lim
n→∞
((1− 1
22
) + ( 1
22
− 1
32
) + . . .+ ( 1
n2
− 1
(n+1)2
)) = 1
2. Determine se a se´rie converge ou se diverge:
(a)
∞∑
n=1
5n3 − 3n
n2(n− 2)(n2 + 5) , (b)
∞∑
n=1
8 arctg n
1 + n2
, (c)
∞∑
n=1
(n+ 3)!
3!n!3n
,
(d)
∞∑
n=1
(−1)n−1 1√
n+ 1
, (e)
∞∑
n=1
(−1)n−1 3 + n
5 + n
.
Soluc¸a˜o. (a) an =
5n3 − 3n
n2(n− 2)(n2 + 5) , bn =
5
n2
lim
n→∞
an
bn
= lim
n→∞
(5n3 − 3n)n2
n2(n− 2)(n2 + 5)5 = limn→∞
5− 3
n2
5(1− 2
n
)(1 + 5
n2
)∑ 5
n2
converge ⇒ ∑ an converge, pelo Teste de Comparac¸a˜o no limite.
(b) an =
8arctg n
1 + n2
≤ bn =
8 pi
2
n2
=
4pi
n2∑ 4pi
n2
converge (p-se´rie com p = 2 > 1)⇒ ∑ an converge, pelo Teste de Comparac¸a˜o.
(c) an =
(n+ 3)!
3!n!3n
> 0 e aplicamos o Teste da Raza˜o.
an+1
an
=
(n+ 4)!
3!(n+ 1)!3n+1
3!n!3n
(n+ 3)!
=
n+ 4
3(n+ 1)
→ 1
3
< 1
A se´rie converge.
(d) Temos
(
1√
x+ 1
)′
= ((x+ 1)−1/2)′ = = −1
2
(x+ 1)−3/2 =
−1
2
√
(x+ 1)3
< 0
e
lim
n→∞
1√
n+ 1
= 0 ⇒ A se´rie converge, pelo Teste da Se´rie Alternada.
3. Determine se a sequeˆncia e´ crescente e se e´ limitada superiormente. Qual e´ o limite
lim
n→∞
an?
(a) an =
3n+ 1
n+ 1
, (b) an =
(2n+ 3)!
(n+ 1)!
, (c) an = 2− 2
n
− 1
2n
.
Soluc¸a˜o. (a) an =
3n+ 1
n+ 1
, e
(
3x+ 1√
x+ 1
)′
=
3(x+ 1)− (3x+ 1)
(x+ 1)2
=
2
(x+ 1)2
> 0
Logo, {an} e´ crescente.
Tambe´m, an ≤ 3n+ 3
n+ 1
=
3(n+ 1)
n+ 1
= 3 e lim
n→∞
an = lim
n→∞
3 + 1
n
1 + 1
n
= 3.
(b) an =
(2n+ 3)!
(n+ 1)!
, an+1 =
(2n+ 5)!
(n+ 2)!
an ≤ an+1 ⇔ (2n+ 3)!
(n+ 1)!
≤ (2n+ 5)!
(n+ 2)!
⇔ 1 ≤ (2n+ 4)(2n+ 5)
n+ 2
⇔ n+ 2 ≤ 4n2 + 18n+ 20, verdadeiro. Logo {an} e´ crescente. E
lim
n→∞
an = lim
n→∞
(2n+ 3)!
(n+ 1)!
= lim
n→∞
(n+ 2)(n+ 3) . . . (2n+ 3) =∞
Donde, {an} na˜o pode ser limitada superiormente.
(c)
2
n
decrescente ⇒ − 2
n
e´ crescente
1
2n
decrescente ⇒ − 1
2n
e´ crescente an = 2− 2
n
− 1
2n
e´ crescente.
an ≤ 2⇒ {an} e´ limitada superiormente por 2.
E lim
n→∞
an = lim
n→∞
(
2− 2
n
− 1
2n
)
= 2.