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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. 17. 18. 19. 20. 21. 22. 23. 24. 25. ,, 8 (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 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. 9
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