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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.
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