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Past Exam for MATH 2300 - Analytic Geometry and Calculus 2 with Newberry at Colorado (CU)

Exam Information

Material Type:Exam 3
Professor:Newberry
Class:MATH 2300 - Analytic Geometry and Calculus 2
Subject:Mathematics
University:University of Colorado - Boulder
Term:--
Keywords:
  • Continuation
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MATH 2300 - Fall 2008 Exam 3 Solutions 1. (a) 1X k=2 34 k+1 = 34 3 1X k=2 34 k 2 = 2764 11 + 3 4 = 2764 47 = 27112 (b) nX k=0 tan 1(k+1) tan 1(k) = n+1X k=1 tan 1(k) nX k=0 tan 1(k) = tan 1(n+1) tan 1(0) = tan 1(n+1): 1X k=0 tan 1(k+ 1) tan 1(k) = lim n!1 nX k=0 tan 1(k+ 1) tan 1(k) = lim n!1tan 1(n+ 1) = 2: 2. (a) lim k!1 1 1=k = 16= 0: So, 1X k=1 1 1=k diverges by the Divergence Test. (b) Since fkg and flnkg are monotone increasing to in nity, then f 1klnkg is monotone de- creasing to zero. So, by the Alternating Series Test, 1X k=2 ( 1)k klnk converges. Also, x and lnx are continuous functions and nonzero for x> 0: So, 1xlnx is also contin- uous. Furthermore, it is decreasing since x and lnx are increasing. Note that Z 1 2 1 xlnxdx = Z 1 ln2 1 udu = lima!1[lnu] a ln2 = lima!1lna ln(ln 2) =1: So, by the Integral Test, 1X k=2 1 klnk diverges. Thus, 1X k=2 ( 1)k klnk converges conditionally. 3. (a) 1X k=1 4 + ( 1)k ...
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