MATH10560 Midterm: Math10560S183Practice Exam3S18solutions
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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) This is a geometric series of the form. 3n arn 1 = a + ar + ar2 + =(converges to diverges a. Xn=1 if |r| < 1 if |r| 1. (technically we should check if an+1/an is a constant r in order to check this. ) We can identify a by calculating the rst term with a1. Then |r| < 1 so the series converges to. 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} .
