MATH10560 Midterm: Math10560Practice Exam3S17solutions
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2. (a) (a) (b) (b) (c) (c) (d) (d) (e) (e) 4. (a) (a) (b) (b) (c) (c) (d) (d) (e) (e) 6. (a) (a) (b) (b) (c) (c) (d) (d) (e) (e) 8. (a) (a) (b) (b) (c) (c) (d) (d) (e) (e) 10. (a) (b) (b) (c) (c) (d) (d) (e) (e) It"s an alternating series with bn = 1/ n. We have n=2 is decreasing since n + 1 > n and thus bn+1 = 1/ n + 1 < (i) the sequence {bn} . 1/ n = bn for all n 2. (ii) limn bn = limn 1/ n = 0. Xn=2 diverges since it"s by the alternating series test. But the series a p series and p = < 1. (a) converges absolutely. (b) diverges because the terms alternate. Xn=2 (c) diverges even though lim n (d) diverges because lim n ( 1)n+1. 6= 0. (e) does not converge absolutely but does converge conditionally.
