and the sum of its first n terms be denoted by. Convergence: An infinite series is said to be convergent if, a definite unique number., finite.

Transcription

1 INFINITE SERIES Seqece: If a set of real mbers a seqece deoted by * + * Or * + * occr accordig to some defiite rle, the it is called + if is fiite + if is ifiite Series: is called a series ad is deoted by Ifiite Series: If the mber of terms i the series is ifiitely large, the it is called ifiite series ad is deoted by ad the sm of its first terms be deoted by Covergece: A ifiite series is said to be coverget if, a defiite iqe mber Eample: / /, fiite Therefore give series is coverget Divergece: teds to either the the ifiite series is said to be diverget Eample: ( ) Therefore is diverget Oscillatory Series: If teds to more tha oe it either fiite or ifiite, the the ifiite series is said to be oscillatory series Eample: : { Therefore series is oscillatory ( ) ( )

2 ( ) { Properties of ifiite series: The covergece or divergece of a ifiite series remais altered o mltiplicatio of each term by The covergece or divergece of a ifiite series remais altered by additio or removal of a fiite mber of its terms Positive term series: A ifiite series i which all the terms after some particlar term are positive is called a positive term series Geometric Series test: The series a Coverges if b Diverges if c Oscillates fiitely if ad oscillates ifiitely if Proof: Let be the partial sm of Case : ie Therefore the series is coverget Case i: ie Therefore the series is diverget Case ii:, Therefore the series is diverget Case i: ie

3 ( ) { Therefore the series is oscillatory Case ii: ie { Therefore the series is oscillatory Note: If a series i which all the terms are positive is coverget, the series remais coverget eve whe some or all of its terms are egative Itegral Test: A positive term series ( ) ( ) ( ) Where ( ) decreases as icreases, coverges or diverges accordig as the itegral ( ) is fiite or ifiite p-series or Harmoic series test: A positive term series is i) Coverget if ii) Diverget if Proof: Let ( ) ( ) [ ] { { Whe, ( ), - Ths coverges if ad diverges if Theorem: Let be a positive term series If is coverget the

4 Proof: If is coverget the ( ) ( ) Note: Coverse eed ot be always tre ie Eve if coverget, the eed ot be Eample : Hece of is diverget by itegral test Bt is a ecessary coditio bt ot a sfficiet coditio for covergece Eample Test the series for covergece, Soltio: Cosider, ( )- Therefore is diverget by Itegral test Eample Test the series for covergece, Soltio: Let The [ ] Therefore is coverget Compariso test: Let ad be two positive term series If a is coverget b The is also coverget That is if a larger series coverges the smaller also coverge

5 Let ad be two positive term series If c is diverget d The is also diverget Eample That is if a smaller series diverges the larger also diverges Test the series for covergece, Soltio: Let ad log log v Bt is a p-series with Therefore is diverget By compariso test is also diverget Eample Test the series for covergece, Soltio: Let ad v Bt Therefore is a geometric series with is coverget By comparisio test is also coverget Aother form of compariso test is Limit test

6 Statemet: If ad be two positive term series sch that ( ) The ad behave alike That is if coverges the also coverge If diverges the also diverge Eamples Test the series for covergece, ( )( ) Choose the Bt with Therefore is coverget By it test is also coverget Eamples 4 Test the series for covergece, ( ) Soltio: ( ) ( ) ( ) Let ( ) ( ) Bt is diverget By it test is also diverget Eamples 5 Test the series for covergece, Soltio:

7 a b (a b)(a ab b ) a b ab a ab b ( ) Let with Bt is coverget By it test is also coverget Eample 6 Test the series for covergece, Solve Soltio: Let with Bt is coverget By it test is also coverget Eample 7 Test the series for covergece, Soltio: We kow that Let The

8 Bt is coverget By it test is also coverget Eample 8 Test the series for covergece, / Soltio: / [ ] [ ] Let The Bt is coverget By it test is also coverget Eercises Test for covergece of the series 4 5 ( ) 6

9 INFINITE SERIES D Alembert s Ratio Test: If ( ) is a series of positive terms, ad the the series is coverget if, is diverget if ad the test fails if If the test fails, oe shold apply compariso test or the Raabe s test, as give below: Raabe s Test: If is a series of positive terms, ad / ( ) the the series is coverget if, is diverget if ad the test fails if Remark: Ratio test ca be applied whe (i) does ot have the form (ii) th term has etc (iii) th term has ( ) ( ) ect (iv) the mber of factors i merator ad deomiator icrease steadily, e: ( ) Eample : Test for covergece the series + +! +! 4 + 4! >> The give series is of the form Therefore + ( )! +! +! +! 4! + whose th term is! ( )!! ( )! ( )(!)

10 Therefore 0 < Therefore by ratio test, is coverget Eample : Discss the atre of the series >> ( ) Therefore + ( )( ) ( )( ) Now ( ) ( )( ) Therefore ( / ) Therefore by D Alembert s ratio test is coverget if diverget if Ad the test fails if Bt whe, ( ) ( ) is of order / (p > ) ad hece is coverget (whe ) Hece we coclde that is coverget ad diverget if > Eample : Fid the atre of series >> Omittig the first term, the give series ca be writte i the form

11 + + + so that Therefore + ( / ) ( / / ) That is, Hece by ratio test is coverget if diverget if ad the test fails if Bt whe, is of order (p > ) Therefore is coverget if ad diverget if > Eample: Fid the atre of the series >> omittig the first term, the geeral term of the series is give by ( ) Therefore + ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( / ) ( / ) ( / )

12 coverget if Hece by ratio test is diverget if ad the fails if Whe, () ( ) ( ) is of order / / (p / > ) ad hece is coverget Therefore is coverget if ad diverget if > Eample : Discs the covergece of the series ( > 0) >> We shall write the give series i the form Now, omittig the first term we have 5( ) [( ) ] 46( ) ( ) ( ) That is, + 5( ) 46( ) That is, + 5( )( ) 46()( ) Therefore 5( )( ) 46()( ) 46 5( ) That is, ( )( ) ( )( )

13 Therefore ( / ) ( / ) ( / ) ( / ) coverget if Hece by ratio test, is diverget if Ad the test fails if Whe, ( )( ) ( )( ) ad we shall apply Raabe s test ( )( ) ( ) (4 0 6) (4 ( ) 4 ) 6 5 ( ) (6 5 / ) ( / ) 6 > 4 Therefore is coverget (whe ) by Rabbe s test Hece we coclde that, is coverget if ad diverget if > Eample : Eamie the covergece of >> Therefore ( / ) ( / ) ( / ( / ) )

14 ( / ( / ) ) ( / ( / ) ) Therefore ( 0) ( 0) ( 0) ( 0) Therefore by ratio test is coverget if ad the test fails if diverget if Whe, Therefore ( / ( / ) ) Sice 0, is diverget (whe ) Hece is coverget if < ad diverget if Eample : test for covergece of the ifiite series!! 4! >> the first term of the give series ca be writte as!/ so that we have,! ( )! ad + ( ) ( )(!) ( )! ( ) Therefore! ( )! ( ) ( / ) ( / ) e < Hece by ratio test is coverget Cachy s Root Test: If is a series of positive terms, ad

15 ( ) ( ), the, the series coverges if, diverges if ad fails if Remark: Root test is sefl whe the terms of the series are of the form, ( )- ( ) We ca ote : (i) (ii) ( ) (iii) ) ( ) Eample : Test for covergece / >> / Therefore ( ) / / / / ( ) / e < Therefore as, also Therefore by Cachy s root test, is coverget

16 Eample : Test for covergece >> Therefore ( ) / / ( ) / e - 4 Therefore e That is, ( ) / e <, therefore e 7 Hece by Cachy s root test, is coverget Eample : Fid the atre of the series / >> / Therefore ( ) / / / / ( ) /

17 e <, sice as, also Therefore by Cachy s root rest, is coverget Eample : Test for covergece >> Therefore ( ) / / ( ) / e -, sice e That is, ( ) / e <, sice e 7 Hece by Cachy s root test, is coverget

18 ALTERNATING SERIES A series i which the terms are alteratively positive or egative is called a alteratig series ie, 4 LEBINITZ S SERIES 4 coverges if A alteratig series (i) each term is merically less tha its precedig term (ii) 0 Note: If 0 the the give series is oscillatory Q Test the covergece of Soltio: Here 7 the + 7( ) 7 6 therefore, (7 6) (7 ) (7 )(7 6) 7 (7 )(7 6) > 0 That is, + > 0, > + Also, 7 (7 / ) 0 Therefore by Leibitz test the give alteratig series is coverget Q Fid the atre of the series log log log 4 log 5

19 Soltio: Here - log( ) the + - log( ) Therefore, + log( ) - log( ) log( ) log( ) log( ) log( ) < 0 Sice ( + ) < ( + ) - + < 0 < + frther - log( ) 0 0 Both the coditios of the Leibitz test are ot satisfied So, we coclde that the series oscillates betwee - ad + Problems: Test the covergece of the followig series i ii ( iii) 4 log log log 4 log5 v iv for 0 ABSOLUTELY & CONDITIONALLY CONVERGENT SERIES A alteratig series a is said to be absoltely coverget if the positive series a a a a4 a is coverget A alteratig series a is said to be coditioally coverget if

20 (i) a is diverget is coverget (ii) a Theorem: A absoltely coverget series is coverget The coverse eed ot be tre be a absoltely coverget series the a Proof: Let a coverget We kow, a a a a4 a a a a4 is By compariso test, a is coverget Q Show that each of the followig series also coverges absoltely (i) a a ; (ii) a ; (iii) a a Soltio: (i) Sice a coverges, we have a 0 as Hece for some positive iteger N, a < for all N This gives a a for all N As a is coverget it follows a coverges (as a is a positive termed series, covergece ad absolte covergece are idetical) (ii) As + a a for all, we get a a the covergece of a a implies the covergece of a (iii) a a a a < a a As a coverges, a 0 as Hece for some positive iteger N, we have a < ½ for all N This gives a a < a for all N Now, by compariso test, a a coverges

21 a That is, a coverges absoltely Q Test the covergece Soltio: Here a the 0 ie, & 0 Ths by Lebiitz rle, a is coverget Also, a Take v The Sice is a 0 v v diverget, therefore a Ths the give series is coditioally coverget is also diverget POWER SERIES A series of the form a a a a i where the a i s are 0 ( ) idepedet of, is called a power series i Sch a series may coverge for some or all vales of INTERVAL OF CONVERGENCE I the power series (i) we have a a Therefore, a a If l, the by ratio test, the series (i) coverges whe a other vales ad diverges for l

22 Ths the power series (i) has a iterval withi which it coverges ad diverges l l for vales of otside the iterval Sch iterval is called the iterval of covergece of the power series Q Fid the iterval of covergece of the series Soltio: Here ad Therefore, By Ratio test the give series coverges for ad diverges for Whe the series redces to coverget, which is a alteratig series ad is 4 Whe - the series becomes, which is diverget (by compariso with 4 p-series whe p) Hece the iterval of covergece is Q Show that the series ( ) is absoltely coverget for <, coditioally coverget for ad diverget for - Soltio Here (-) - Therefore + ( ) N ( ) ( ) ( ) ( / ) ( ) ( / ) Therefore by geeralized D Alembert s test the series is absoltely coverget if <, ot coverget if > ad the test fails if

23 Now for, ca be + or If the give series becomes Here, + Bt + < + > + Also Therefore by Leibitz test the series is covergece whe Bt the absolte series order / whose geeral term is 5 7 ad hece is diverget ad is of Sice the alteratig series is coverget ad the absolte series is diverget whe, the series is coditioally coverget whe If -, the series becomes where the series of positive terms is diverget as show already 5 7 Therefore the give series is diverget whe - Ths we have established all the reslts Problems: Test the coditioal covergece of ( i) ii si si si Prove that is absoltely coverget For what vales of the followig series are coverget 4 i 4 4 ii 4 4 iii iv

24 4 Test the atre of covergece ******