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by: Vivien Bradtke V


Vivien Bradtke V
Texas A&M
GPA 3.6


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Class Notes
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This 5 page Class Notes was uploaded by Vivien Bradtke V on Wednesday October 21, 2015. The Class Notes belongs to MATH 152 at Texas A&M University taught by Staff in Fall. Since its upload, it has received 13 views. For similar materials see /class/226040/math-152-texas-a-m-university in Mathematics (M) at Texas A&M University.

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Date Created: 10/21/15
Math 152 honors sections Harold P Boas b0astamuedu Reminder Second examination is Wednesday November 3 The exam covers through section 104 An old exam is posted at http www math tamu eduboascourses 152 2002cexam2 pdf Math 152 October 29 2004 i slide 2 Convergence tests so far 00 If an 74gt 0 then 2 an diverges 711 A geometric series with lratiol lt 1 converges DO Integral test for positive decreasing functions the improper integral f x dx and the 1 00 corresponding series 2 f either both converge or both diverge 711 Inequality comparison test for positive terms if 0 lt an lt bn at least for 11 large and if 2 bn converges then 2 an converges too 711 711 11 GO in exists finite limit and if 2 bn con 71 711 Limit comparison test for positive terms if lim b 7100 00 verges then 2 an converges too 711 Math 152 October 29 2004 i slide 3 Root test not in book Example 2 211 converges n 1 This is not a geometric series and it is bigger than the convergent geometric series 1 g so the n comparison test does not seem to help 1n n n n n n 11 i 1n i 7 Since 2 72 gt and since n 1 we have 2 lt lt 2 gt when n is large so we 11 n can use the comparison test after all compare to the convergent geometric series 2 Math 152 October 29 2004 i slide 4 Root test and ratio test 00 Root test If 0 lt an and if lim LIEn lt 1 then 2 an converges Moreover if lim 5111 quot gt 1 then ngtoo 1 ngtoo 00 2 an diverges because then an 74gt 0 If lim bign 1 the test gives no information 711 naoo Ratio test Exactly the same as the root test except look at lim 21 instead of lim 5111 quot HHOO n HHOO oo 3nltnl2 Example 5 lt2 n1 I I n I I 2 hm unH hm 3 n 1n 1 3 hm 3n 1 E lt 1 H00 an ngtoo ngt00 4 so the original series converges Math 152 October 29 2004 i slide 5 Series with some negative terms Negative terms can only help with convergence 00 00 if E qu converges then so does 2 an n An absolutely convergent series converges 0 7 n 00 Example E 12 converges because 2 converges n1 n1 lt71quot 7712 1 712 The two series sum to different values however 2 7 and E i 7 nil 2 2 nil 2 6 1 Example 2 a diverges harmonic series 1 lt71 1 yet 2 n converges in fact to the value ln n1 Math 152 October 29 2004 i slide 6 Alternating series test If an i 0 then X 71 51n converges n 1 Error estimate When the alternating series test applies the sum of the series is trapped be tween any two consecutive partial sums 71quot 1 1 1 1 Example271g7 iau 1111 71quot 111 1 iii iii 1 iii f 24 3444 54ltn i 4 lt 24 3444 01 7094754lt E 4 lt709459 n1 n lt71 4 Exact Value 2 T 7275 m 70947033 n1 Math 152 October 29 2004 i slide 7 Homework 0 Read section 104 pages 605 610 0 Do the Suggested Homework problems for section 104 Monday we will review for the exam and look at an old exam Math 152 October 29 2004 i slide 8


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