MTH 253 Calculus (Other Topics)Does the Series Converge?Alternating SeriesThe Alternating Series TestSlide 5Example of the Alternating Series TestApproximating Alternating SeriesSlide 8Slide 9Absolute ConvergenceAbsolute Convergence: Example 1Absolute Convergence: Example 2Absolute Convergence TheoremAbsolute Convergence Theorem: ExampleConditional ConvergenceRatio Test for Absolute ConvergenceMTH 253Calculus (Other Topics)Chapter 10 – Infinite SeriesSection 10.6 – Alternating Series; Conditional ConvergenceCopyright © 2006 by Ron Wallace, all rights reserved.Does the Series Converge?1kku�=�8 Tests for ConvergenceDivergence TestIntegral Testp-Series TestComparison TestLimit Comparison TestRatio TestRoot TestAlternating Series TestEach test has it limitations (i.e. conditions where the test fails).The test tells you nothing!10.410.510.6Series with positive terms!Alternating Series11 2 3 4( 1) kka a a a a+- = - + - +�ggg1 2 3 4( 1) kka a a a a- =- + - + -�gggORNOTE: All ak’s are assumed to be positive.The Alternating Series Test11 2 3 4( 1) kka a a a a+- = - + - +�ggg1 2 3 4( 1) kka a a a a- =- + - + -�gggORConverges if …11. k ka a k+� "k2. lim 0ka��=The Alternating Series Test11 2 3 4( 1) kka a a a a+- = - + - +�gggConverges if …11. k ka a k+� "k2. lim 0ka��=2 12 1increasing and boundeddecreasing and boundedlim lim 0 nnnns as��-��<>{ }{ }22 1n En Os Ss S-��2 1 2 2Since: n n ns s a-= +2 1 2 2lim lim( ) 0O n n n E En nS s s a S S-�� ��= = + = + =Therefore, under these conditions, the alternating series converges.Proof:Example of the Alternating Series Test1ln( 1)kkk+-�Decreasing?2 2ln (1/ ) ln 1 ln0 when 3d x x x x xxdx x x x- -� �= = < �� �� �Limit?ln 1/lim lim 01k kk kk�� ��= =Therefore, convergent.Approximating Alternating SeriesIf an alternating series satisfies the conditions of the alternating series test, and sn is used to approximate the sum; then …1n nS s a+- <i.e. The error is less than the first term omitted.Approximating Alternating SeriesExample:1. Estimate the error if 4 terms are used to approximate the sum.11( 1)!kkk+�=-�5 1( 1) 1.0083335! 120+-= �Approximating Alternating SeriesExample:11( 1)!kkk+�=-�2. How many terms are need to make sure the error is less than 0.01?( 1) 1( 1)0.01( 1)!nn+ +-<+1.0416664!1.0083335!��10.01( 1)!n<+Therefore, four terms are needed!Absolute Convergenceku�converges absolutely if …ku�is convergent.Otherwise, it diverges absolutely.Absolute Convergence: Example 11 1 1 1 1 1 112 4 8 16 32 64 128ku = - - + + - - + +�ggg112kku-=� �Convergent geometric series, therefore the first series converges absolutely.Absolute Convergence: Example 21 1 1 1 1 1 112 3 4 6 7 8 9ku = - + - + - + - +�ggg1kuk=� �Divergent harmonic series, therefore the first series diverges absolutely.NOTE: The first series IS a CONVERGENT alternating series.Absolute Convergence TheoremIf a series converges absolutely, then it converges.Proof:( ) ( )k k k k k k ku u u u u u u� �= + - = + -� �� � � �convergent0 or 22 2k k kk k k ku u uu u u u+ =\ + � =� � �convergentTherefore, by the comparison test, the series converges.Absolute Convergence Theorem: Example1 1 1 1 1 1 112 4 8 16 32 64 128ku = - - + + - - + +�ggg112kku-=� �Convergent geometric series, therefore the first series converges absolutely.Therefore, the original series converges.NOTE: If it does not converge absolutely, the test fails!Conditional ConvergenceIf a convergent series diverges absolutely, it is said to be conditionally convergent.Example:11( 1)kk+-�Convergent alternating series.11 1( 1)kk k+- =� �Divergent harmonic series.Therefore, the first series is “Conditionally Convergent.”Ratio Test for Absolute Convergence1Let: limkkkuLu+��=Given ku�• If L < 1, the series converges absolutely.• If L > 1, the series diverges.• If L = 1, the test fails.NOTE: A summary of all of the convergence tests is given on page