s k 1 4 k 2 3 k B s n 2 3 1 n 2 n n are convergent 1 A only 2 both of them 3 B

S k 1 4 k 2 3 k b s n 2 3 1 n 2 n n are convergent 1

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s k =1 4 k 2 + 3 k B. s n = 2 3 + ( - 1) n 2 + n n are convergent. 1. A only 2. both of them 3. B only correct 4. neither of them Explanation: A. Divergent: use Limit Comparison Test and Geometric Series. B. Convergent: use Limit Comparison Test and p -series Test with p = 3 2 . 016 10.0 points Determine whether the series s k =1 sin p 2 k P is absolutely convergent, conditionally con- vergent or divergent. 1. conditionally convergent 2. absolutely convergent 3. divergent correct Explanation: All the terms in the series s k =1 sin p 2 k P are positive for k 2, so the given series ei- ther converges absolutely or it is divergent.
choice (hac762) – HW Quest Week 8 – cepparo – (53850) 8 To determine which we use the Limit Com- parison Test with a k = sin p 2 k P , b k = 2 k . For then lim k →∞ a k b k = lim k →∞ sin p 2 k P 2 k = lim θ 0 sin( θ ) θ = 1 > 0 . Thus the given series s k =1 sin p 2 k P converges if and only if the series s k = 1 2 k converges. But by the p -series test with p = 1 (or because the harmonic series is divergent), this last series is divergent. Consequently, the given series is divergent . 017 10.0 points Compare the radius of convergence, R 1 , of the series s n =0 c n x n with the radius of convergence, R 2 , of the series s n = 1 n c n x n - 1 when lim n →∞ v v v c n +1 c n v v v = 3 . 1. 2 R 1 = R 2 = 3 2. R 1 = 2 R 2 = 3 3. 2 R 1 = R 2 = 1 3 4. R 1 = 2 R 2 = 1 3 5. R 1 = R 2 = 3 6. R 1 = R 2 = 1 3 correct Explanation: When lim n →∞ v v v c n +1 c n v v v = 3 , the Ratio Test ensures that the series s n =0 c n x n is (i) convergent when | x | < 1 3 , and (ii) divergent when | x | > 1 3 . On the other hand, since lim n →∞ v v v ( n + 1) c n +1 nc n v v v = lim n →∞ v v v c n +1 c n v v v , the Ratio Test ensures also that the series s n =1 n c n x n - 1 is (i) convergent when | x | < 1 3 , and (ii) divergent when | x | > 1 3 . Consequently, R 1 = R 2 = 1 3 . 018 10.0 points
choice (hac762) – HW Quest Week 8 – cepparo – (53850) 9 Find a power series representation for the function f ( x ) = 1 x - 2 . 1. f ( x ) = - s n =0 1 2 n +1 x n correct 2. f ( x ) = s n =0 ( - 1) n 2 n x n 3. f ( x ) = s n =0 ( - 1) n - 1 2 n +1 x n 4. f ( x ) = - s n =0 2 n x n 5. f ( x ) = s n =0 1 2 n +1 x n Explanation: We know that 1 1 - x = 1 + x + x 2 + . . . = s n = 0 x n . On the other hand, 1 x - 2 = - 1 2 p 1 1 - ( x/ 2) P . Thus f ( x ) = - 1 2 s n = 0 p x 2 P n = - 1 2 s n = 0 1 2 n x n . Consequently, f ( x ) = - s n = 0 1 2 n +1 x n with | x | < 2. 019 10.0 points Find a power series representation for the function f ( z ) = 1 4 - z 3 . 1. f ( z ) = s n =0 z 3 n 4 3 n 2. f ( z ) = - s n =0 4 n z 3 n 3. f ( z ) = - s n =0 z n 4 n +1 4. f ( z ) = s n =0 4 n z 3 n 5. f ( z ) = s n =0 z 3 n 4 n +1 correct 6. f ( z ) = - s n =0 z 3 n 4 3 n Explanation: After simpli±cation, f ( z ) = 1 4 - z 3 = 1 4 p 1 1 - ( z 3 / 4) P . On the other hand, we know that 1 1 - t = s n =0 t n . Replacing t with z 3 / 4, we thus obtain f ( z ) = 1 4 s n =0 z 3 n 4 n = s n =0 z 3 n 4 n +1 .

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