SUTCLIFFE S NOTES: CALCULUS 2 SWOKOWSKI S CHAPTER 11

Transcription

1 SUTCLIFFE S NOTES: CALCULUS SWOKOWSKI S CHAPTER Ifiite Series.5 Altertig Series d Absolute Covergece Next, let us cosider series with both positive d egtive terms. The simplest d most useful is ltertig series, i which the terms re ltertely positive d egtive. with 0 for every. or I the followig discussio we will cosider oly the first cse, mely, Theorem (Altertig Series Test) The ltertig series 3 4 is coverget if the followig two coditios re stisfied: Proof: (i) 0 for every (ii) lim 0 By (i) we my write Cosider the prtil sums 3 4. S, S, S,, S, 4 6 which coti eve umber of terms of the series. Sice S d 0 for every, the ; S S4 S tht is, S is mootoic sequece.

2 Refer to Figure. o pge 558. We see tht S for every positive iteger. We c lso prove this lgebriclly by observig tht Thus, S S is bouded mootoic sequece d there exists umber S such tht lims S. We ext cosider prtil sum S hvig odd umber of terms of the series. The d sice by (ii) lim 0, S S lim S lim S S. Sice both the sequece of eve prtil sums d the sequece of odd prtil sums hve the sme limit S, it follows tht i.e. the series coverges. Exercises lims S ; Determie whether the series () stisfies the coditios of the ltertig series test (b) coverges or diverges # 5 Solutio: There re severl wys to prove (i). is decresig for x. Method Show tht f x x5 x x x x f x 5 x5 l5 5 xl5 0 for x Method Direct proof: show Method 3 Show tht

3 To prove (ii): LH ' lim 5 0 lim lim l5 Note: The AST my lso be used if coditio (i) holds for m for some positive iteger m, becuse this correspods to deletig the first m terms of the series. If series coverges, the the th prtil sums c be used to pproximte the sum S of the series. For ltertig series, we c pproximte the error i the pproximtio s follows: Theorem Let the sum of the series d be ltertig series tht stisfies coditios (i) d (ii) of the AST. If S is S is prtil sum, the S S ; tht is, the error ivolved i pproximtig S by S is less th or equl to. Proof: The series obtied by deletig the first terms of 3 3, lso stisfies the coditios of the AST d therefore hs sum S S R 3 R. d 3, mely, R. Thus, Usig the sme rgumet we used i the proof of the AST, we see tht R. Cosequetly, E S S R which is wht we wished to prove. The Accurcy of Approximtio -deciml ccurcy if -deciml ccurcy if 3-deciml ccurcy if E E E

4 I the followig, pproximte the sum to 3 deciml plces. #34 0! Solutio: lim 0!!!! Thus, the series coverges by the AST ! 000 4! ! 0!! 4! 6! 8! So we see tht ! Hece, the prtil sum S 4 pproximtes S to 3 deciml plces. S S The followig is useful for ivestigtig series cotiig both positive d egtive terms but is ot ltertig. Defiitio A series is coverget. is bsolutely coverget if the series Exmple 3 Prove tht the ltertig series coverget. is bsolutely 3 4 Proof. Tig the bsolute vlue of ech term gives us the coverget p-series (p=). Exmple 4 The ltertig hrmoic series is 3 4 Show tht this series is () coverget (b) ot bsolutely coverget Solutio: () Coditios (i) d (ii) of the AST re stisfied: for every d lim 0. Hece, the ltertig hrmoic series is coverget. 4

5 This is the diverget hrmoic series. Thus, the series is ot bsolutely coverget. (b) Defiitio A series Theorem If series is coditiolly coverget if is bsolutely coverget, the Proof: If we let b d use the property, the If 0, or 0 b. is bsolutely coverget, the pply the BCT, it follows tht the proof is complete. is coverget d is diverget. is coverget. is coverget d hece, is coverget. If we b is coverget. Sice b, b is coverget. The Exmple 5 p. 563 ot ltertig or geometric or positive-term coverget geometric series The origil series is bsolutely coverget d thus coverget. Exmple 6 si si 3 si si cotis both positive d egtive terms but is ot ltertig 3 si si (coverget p series) is coverget Thus, is bsolutely coverget d therefore is coverget. The 3 Clssifictios of Series (i) bsolutely coverget (ii) coditiolly coverget (iii) diverget Theorem (Rtio Test for Absolute Covergece) Let be series of ozero terms, d suppose (i) If L, the series is bsolutely coverget. (ii) If or lim L, the series is diverget. (iii) If L, use differet test. lim L. Note: There is lso root test for bsolute covergece; replce by. 5

6 #4 Solutio:! 5! 5 lim lim lim 5! 5 #4 Solutio: #6 Solutio:! 5! is diverget! lim lim lim lim lim 0! is bsolutely coverget! lim lim lim diverges 6