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Calculus II Exam 3, Fall 2002, Answers 1. Find the limits ln x 1 a) lim x e ln ln x 1 x x e 1 1 ln x x Answer. l H lim x 1 2x b) lim x ∞ 3x2 1 Answer. lim lnx ln e 1 x e x 2x2 ∞ 3x2 1 lim x lim x ∞ 2 3 1 x 1 x2 2 3 2. Does the integral converge or diverge? Give reasons. If you can, evaluate the integral. 1 a) 0 dx x9 10 Answer. ∞ b) 0 1 lim a 0 a x 9 10 dx lim 10x1 a 0 10 1 a lim 10 1 a1 a 0 10 10 x dx 1 x3 Answer. This converges because x 1 x3 and 1 x2 ∞ 0 1 x 1 x2 dx ∞ x2 3. Does the series converge or diverge? Give reasons. ∞ a) n 2n 1 ∑ 3n n 0 Answer. This series converges by comparison with ∞ 2 n 0 3 ∑n n which is 2 3 times the derived geometric series. Here is the comparison: 2 2 n 2n 1 n n n n n 3n 3 3 3 ∞ b. n 1 ∑ ln n n 0 Answer. This diverges by comparison with the series ∑ n 1 , since ln n n. 4. What is the radius of convergence of the power series? Show your work. ∞ a) ∑ n 0 2n 1 xn Answer. We calculate the ratios of the coefficients: an 1 an 1 2n 1 1 2 n 2 1 1 1 2n 1 1 2n 2 1 R so R 1 2. ∞ b) 3n ∑ n! xn n 0 Answer. We calculate the ratios of the coefficients: an 1 an 3n 1 n! n 1 ! 3n 3 0 n 1 so R ∞. 5. Find the Maclaurin series for the function. 1 x 1 x a) Answer. We start with the geometric series: ∞ 1 ∑ xn 1 x n 0 and then multiply by 1 x: 1 x 1 x 1 1 x x 1 x x b) 0 ∞ ∞ ∑ xn x ∑ xn n 0 ∞ ∞ n 0 n 1 n 0 ∞ ∑ xn ∑ xn n 1 Answer. We start with the geometric series: ∞ ∑ xn 1 x n 0 and substitute t 3 for x: 1 1 t3 ∞ ∑ t 3n n 0 and finally, integrate (term by term on the right): x 0 dt 1 t3 ∞ n 0 n 0 1 2 ∑ xn dt 1 t3 1 ∞ ∑ xn ∑ xn ∞ t 3n ∑ 3n n 0 1 1 1