MTH 161 Midterm: MTH 161 University of Rochester 161 Midterm 1 Fall 2006

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31 Jan 2019
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Express the following continued fraction as a rational number. If you can express it as the sum of two rational numbers, that"s also ok. Espressing the number in terms of a geometric series, we nd. This is a telescoping series, and all but three terms cancel out, so. If we look at the dominant terms, we get 2n/ n2 = 2, so we suspect that the series diverges by the divergence test, since the limit of the terms is not 0. To make this rigorous, we divide top and bottom by the leading power of n, namely just n, to get. So, the series diverges by the divergence test: (11 points) The series converges. (b) (6 points) justify your answer in part (a). This series can be analyzed using the integral test. Using the substitution u = ln x, du = 2 and so the series converges: (12 points)

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