MTH 253 Calculus (Other Topics)Does the Series Converge?The Comparison TestUsing the Comparison TestExamples w/ the Comparison TestThe Limit Comparison TestExample w/ the Limit Comparison TestThe Ratio TestSlide 9Slide 10Slide 11Example w/ the Ratio TestThe Root TestExample w/ the Root TestMTH 253Calculus (Other Topics)Chapter 10 – Infinite SeriesSection 10.5 – The Comparison, Ratio, and Root TestsCopyright © 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!10.410.5The Comparison TestBasic Idea:If you can show that a first series is less than a second series, and the second series is know to be convergent, then so is the first series (sn increasing w/ upper bound).If you can show that a first series is greater than a second series, and the second series is divergent, then so is the first series (sn increasing wo/ upper bound).k ka b�� �k ka b�� �Difficult Part? Finding the series to compare.Using the Comparison Test1. Make an “educated” guess.•convergent or divergent?2. Find a series and prove your guess.•usually similar to the originalHelpful Ideas:Increasing the numerator or decreasing the denominator gives something bigger.Decreasing the numerator or increasing the denominator gives something smaller.Examples w/ the Comparison Test19 6k +�1 1 19 6 9k k k< <+Smaller than the divergent harmonic series NOTHING!42k k+�4 4 42 2 12k k k k< = �+Smaller than twice a convergent p-series Convergent21kk k+-�2 2 21 1k k kk k k k k k+> > =- -Larger than the divergent harmonic series DivergentThe Limit Comparison TestBasic Idea:If two series essentially differ by a constant (except for possibly the first finite number of terms), then they have the same behavior.lim 0 and finitekkkab��>Difficult Part? Finding the series to compare.Example w/ the Limit Comparison Test19 6k +�1 1 19 6 9k k k< <+Smaller than the divergent harmonic series NOTHING!Comparison Test Fails!1 9 6lim lim 91 9 6k kk kk k�� ��+= =+Therefore, same behavior Divergent!The Ratio Test1Let: limkkkuLu+��=�1 where kkur r L k Kue+< > + " >21 1 1...kk k ku ru r u r u+ -< < <1 1 kku r u+<\� ��Geometric Series!Convergent if |r|<1 Comparison Test If L < 1, the series converges.The Ratio Test1Let: limkkkuLu+��=�1 where kkus s L k Kue+> < + " >21 1 1...kk k ku su s u s u+ -> > >1 1 kku s u+>\� ��Geometric Series!Divergent if |s|>1 Comparison Test If L > 1, the series diverges.The Ratio Test1Let: lim 1kkkuLu+��= =If L=1, the test fails!1 divergesk�1 ( 1)lim lim 11 1k kk kk k�� ��+= =+21 convergesk�2 22 21 ( 1)lim lim 11 2 1k kk kk k k�� ��+= =+ +The Ratio Test1Let: limkkkuLu+��=Given where 0 k ku u k> "�• If L < 1, the series converges.• If L > 1, the series diverges.• If L = 1, the test fails.Example w/ the Ratio Test11 2 222 2244 4( 1)lim lim lim 4 14( 1) 4 2 1kkkkk k kk kkk k kk++�� �� ��+= � = = >+ + +24kk� Divergent!The Root TestLet: limkkku L��=Given where 0 k ku u k> "�• If L < 1, the series converges.• If L > 1, the series diverges.• If L = 1, the test fails.Proof is similar to the ratio test!Example w/ the Root Test1 1 1lim lim 12 2 2kk kkk ke e- -�� ��� �- -= = <� �� �12kke-� �-� �� ��
View Full Document