# convergence tests and taylor series-solutions - lee(jsl2382...

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lee (jsl2382) – convergence tests and taylor series – tran – (56320) 1 This print-out should have 50 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. This homework includes convergence tests (oF all kinds), and Taylor series. 001 10.0 points Determine the interval oF convergence oF the series s n = 1 ( 2) n n + 4 x n . 1. p 1 2 , 1 2 b correct 2. ( 2 , 2] 3. ( −∞ , ) 4. [ 4 , 4) 5. converges only at x = 0 6. B 1 4 , 1 4 P 7. none oF the other answers Explanation: The given series has the Form s n = 1 a n x n with a n = ( 2) n n + 4 . Now For this series, (i) R = 0 if it converges only at x = 0, (ii) R = if it converges for all x , while iF R > 0, (iii) it converges when | x | < R , and (iv) diverges when | x | > R . But lim n →∞ v v v a n +1 a n v v v = lim n 2( n + 4) n + 5 = 2 . By the Ratio Test, thereFore, the given series converges when | x | < 1 / 2 and diverges when | x | > 1 / 2. On the other hand, at the points x = 1 / 2 and x = 1 / 2 the series reduces to s n = 1 1 n + 4 , s n = 1 ( 1) n n + 4 respectively. Now, by the p -series Test with p = 1, the frst oF these series diverges, while by the Alternating Series Test, the second conveges. Consequently, the given series has interval oF convergence = p 1 2 , 1 2 b . keywords: 002 10.0 points Determine the convergence or divergence oF the series ( A ) s k = 1 3 ln(2 k ) k 2 , and ( B ) s k = 1 sin 2 k k 2 + 4 . 1. A converges, B diverges 2. both series diverge 3. both series converge correct 4. A diverges, B converges Explanation: ( A ) The Function f ( x ) = 3 ln 2 x x 2
lee (jsl2382) – convergence tests and taylor series – tran – (56320) 2 is continuous and positive on [1 , ); in addi- tion, since f ( x ) = 3 p 1 2 ln2 x x 3 P < 0 on [1 , ), f is also decreasing on this inter- val. This suggests applying the Integral Test. Now, after Integration by Parts, we see that i t 1 f ( x ) dx = 3 b ln(2 x ) x 1 x B t 1 , and so i 1 f ( x ) dx = 3(1 + ln 2) . The Integral Test thus ensures that series ( A ) converges . ( B ) Note Frst that the inequalities 0 < sin 2 k k 2 + 4 1 k 2 + 4 1 k 2 hold for all n 1. On the other hand, by the p -series test the series s k = 1 1 k 2 is convergent since p = 2 > 1. Thus, by the comparison test, series ( B ) converges . 003 10.0 points To apply the root test to an inFnite series s k a k , the value of ρ = lim k →∞ | a k | 1 /k has to be determined. Compute the value of ρ for the series s k = 1 2 k k (ln k + 5) k . 1. ρ = 2. ρ = 0 correct 3. ρ = 2 4. ρ = 10 5. ρ = 5 Explanation: ±or the given series ( a k ) 1 /k = 2 1 /k p ln k + 5 k P = 2 1 /k p ln k k + 5 k P . But 2 1 /k −→ 1 , ln k k 0 as k → ∞ . Consequently, ρ = 0 . 004 10.0 points Which of the following series ( A ) s n = 3 5 n + 7 ( n ln n ) 2 + 6 ( B ) s n = 1 n 8 n + 7 ( C ) s n = 1 p 8 n + 7 6 n 5 P n diverge(s)? 1. B and C correct 2. C only 3. all of A, B, C 4. A and B
lee (jsl2382) – convergence tests and taylor series – tran – (56320) 3 5. B only Explanation: ( A ) After division, 5 n + 7 ( n ln n ) 2 + 6 = 5 + 7 n n (ln n ) 2 + 6 n .