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M1220-2 Quiz 9 - Homework Quiz Spring 2005 Quiz Scores (out of 10): n = 121; mean = 7.8 25th percentile = 7 ; median (50th percentile) = 8 ; 75th percentile = 9.5 1. (6 pts) State whether the following series are absolutely convergent, conditionally convergent, or divergent. Justify your answer by listing the test used and showing the test results. i. +∞ X n(−3)n n=1 4n−1 n+1 Note that limn→+∞ |an+1 /an | = limn→+∞ (n+1)3 4n by the Ratio Test, this series converges absolutely. ii. 4n−1 n3n = limn→+∞ n+1 3 n 4 = 3 4 < 1. So, +∞ X n! (−5)n n=1 (n+1)! 5n (n+1)! 5n 5n+1 n! = limn→+∞ n! 5n+1 P+∞ n! Test, n=1 (−5)n diverges. Note that limn→+∞ |an+1 /an | = limn→+∞ limn→+∞ n+1 5 = +∞. So, by the Ratio iii. +∞ X n=1 (−1)n = n 3 + n2 Note that limn→+∞ |an+1 /an | = limn→+∞ n+1 3+n2 3+(n+1)2 n =1 P n P so the Ratio Test is inclusive. However, the series, 1/n 3+n2 is similar to the series, which diverges. Since h i n limn→+∞ 3+n2 / n1 = 1, P n by the Limit Comparison Test, the series 3+n2 also diverges. In that case, the given series cannot converge absolutely. However, since the terms of the given series, ignoring their sign, are decreasing and since P+∞ n n n limn→+∞ 3+n 2 = 0, by the Alternating Series Test, the series n=1 (−1) 3+n2 converges conditionally. 2. (2 pts) Write out the first 4 terms in the power series given below. Then find the convergence set for the power series and the radius of convergence. +∞ X (−3)n xn √ n+1 n=o The first four terms are: 1 − 3x √ 2 + 2 9x √ 3 n+1 − n+1 27x3 √ 4 √ √ n+1 x √n+1 limn→+∞ |an+1 /an | = limn→+∞ | 3 √n+2 3n xn | = limn→+∞ |3x n+2 | = |3x|. But |3x| < 1when −1/3 < x < 1/3 so the power series converges when −1/3 < x < 1/3. At the P (−1)n √ endpoints of the interval: when x = 1/3, the above series becomes which converges n+1 P 1 √ by the Alternating Series Test; when x = −1/3, the above series becomes which n+1 diverges as it is a p-series with p ≤ 1. So, the convergence set for this series is (−1/3, 1/3] and the radius of convergence is R = 1/3. 3. (2 pts) Write out the first 4 terms in the power series given below. Then find the convergence set for the power series and the radius of convergence. +∞ X (x − 2)n nn n=1 The first four terms are: (x − 2) + (x − 2)2 /22 + (x − 2)3 /33 + (x − 2)4 /44 . n+1 n (x−2) n x−2 limn→+∞ |an+1 /an | = limn→+∞ | (n+1) n+1 (x−2)n | = limn→+∞ | n+1 1 n n n n n+1 ( n+1 ) . Since limn→+∞ ( n+1 ) 1 n n limn→+∞ ( n+1 ( n+1 ) = 0. So, the convergence set |x − 2| limn→+∞ radius of convergence is R = +∞. h n n+1 in = 1/e and limn→+∞ |= 1 n+1 = 0, for this series is (−∞, +∞) and the