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by: Jack Magann

134

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4

# Chapters 8.4 cont. and 8.5 MTH 162

Marketplace > University of Miami > Mathematics (M) > MTH 162 > Chapters 8 4 cont and 8 5
Jack Magann
UM
GPA 3.865
Calculus 2
Dr. Bibby

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These notes continue with examples from 8.4 and chapter 8.5. Topics include: examples of the root test, the alternating series test and the power test.
COURSE
Calculus 2
PROF.
Dr. Bibby
TYPE
Class Notes
PAGES
4
WORDS
CONCEPTS
University of Miami Calculus 2
KARMA
25 ?

## Popular in Mathematics (M)

This 4 page Class Notes was uploaded by Jack Magann on Friday March 27, 2015. The Class Notes belongs to MTH 162 at University of Miami taught by Dr. Bibby in Spring2015. Since its upload, it has received 134 views. For similar materials see Calculus 2 in Mathematics (M) at University of Miami.

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Date Created: 03/27/15
Calculus 2 chapter 84 Root Test Examples 2n3 quot n 2n3 quot 2n3 E 1 3n2 L hmquotquot 3n2 hmquotquot 3n2 3 E 2n3 quot Slnce L 3 lt 1 The ser1es Z 3n2 converges 2 Z311273n 5n322n n 311273n 311273 27n9 a L hmquotquot 511322n hmquotquot 511322 hmquotquot 25n6 25 a gt 1 Th 23n273n d t 25 e ser1es 5n322n 1s 1vergen Alternating Series Test for convergence Tl An alternating series can be expressed as Z 1 Let 2 an be an alternating series If 1 an1 lt Ianl and 2 limnnmlanl 0 Then 2 an is convergent Ex 1n1 1 1 2 n25 cons1der2an n25 2 2 1 1 Set up an equality n 5 gt n so n25 n2 X converges pseries p 2 X 1125 converges by the Basic Comparison Test 1n1 X n25 absolutely converges by the alternating ser1es test 1quot1 2 Z n also referred to as the alternating harmonlc ser1es Elanl 2 This diverges pseries p 1 However an1 lt l lanl and limngoolanl limnami O n1 n Conditions one and two are met showing convergence n1 Z 1 is convergent by the alternating series test 1quot1 1s conditlonally convergent Calculus 2 Chapter 85 Power Series 213 0anx C a0 0106 C a2x C2 For39E xn c O a 1 For39 E nlxn c O a n 39 n2n n n2quot 39 1x3x52n 1 n 1x3x52n 1 These series converge on an interval centered at X c called the interval of convergence There is a number R 0 S R S 00 called the radius of convergence 1 If R 0 series converges for x c and diverges for all x i c 2 If R 00 series converges for all X 3 If 0 lt R lt 00 series converges absolutely for Ix cl lt R Endpoints c R and c R must be tested separately by plugging them in for X Interval of convergence Inequlity Interval Behavior at endpoint c RltxltcR c RcR Divergesatboth c RSxltcR c RcR c RconvergescRdiverges c RltxScR c RcR c RdivergescRconverges c R SxScR c RcR Convergesat both Don t use ratio or root test on the endpoints an for Z anx cquot or R limnoo kaa n for Z anx ckquot n1 Proof consider the power series 2 an x cquot Let un anx C un an1x Cn1 By ratio test series converges if un1 an1x Cn1 lt1 r limnoo limnoo anx Cn gt limnoo Z x cquot lt 1 Tl gt limnoo an x cquot lt 1 gt x cquot lt lim aquot R n gtoo an1 Ex xquot 1 23 an n C 0 L 1 R limnoo limnoo quotin limnw x n 12quot2 n1 n2quot n12n1 Emu gt00 n12 2 n Now test endpoints at c R 2 and c R 2 Z i 2 l diverges harmonic series n2quot n 2quot 1quot X n X T converges alternat1ng harmon1c ser1es Tl 2 16 converges absolutely on the interval 2 2 xquot 1 1 2 25 6111 n1m C0 R limnoo limnoo limnoo ixn 1 limnoo n 1 00 am n1 n Series converges on the interval 00 oo 32xn 6111 an C20 n23quot n23quot n123n1 1 n23quot 1 n123n1 R limnm E limnoo limnw nZ lsnxm 123quot3 2 limnoo w 3 test endpoints at 3 and 3 n2 Tl X nisn 2 converges pseries p 2 3quot 1quot lc lquot X n23n 2 cons1der 2 n2 n2 1 I 2 E converges pseries p Z 2 2 11926311 converges on the interval 3 3 nxquot n n1 an 4 an 2 S n an 2 a C 0 R 11mnoo E n 5nx5 5n E R mun 3 5 n n1 mun 3 n1 1 5 1 at x 5 21 limnoo quotSin 2 limnoo n 00 diverges 2 at x 5 Zni ns limnoo n 1quot n 1X1quot1 lt 111quot and limnoo 111quot 2 00 diverges Tl 2 quotSin converges on the interval 5 5 I nlxquot 5 F1nd the rad1us of convergence of E 11 aquot 1x3x5x2n 1 an1 1X3gtlt5gtlt2n1 C 0 R rm 4 1 n gtoo 1X3X5X2n1 n1 I 1x3x5x2n 12n1 llmnqoo 1X3X5Xquot2n1 n1n limn m 2n1 2 n1

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