《高等数学》课程电子教案(PPT课件)Chapter 12 Infinite series Sec 2 Convergence Tests for Series of Constant

:12ChapterInfiniteseriesSec.2Convergence Tests for Seriesof Constant$ 10.2
Chapter 12 Infinite series Sec.2 Convergence Tests for Series of Constant §10.2

$ 12.2 Convergence tests for series of constantThere arealways two important questionsto askaboutaseries.1.Doestheseries converge?2. If it converges, what is its sum?
§12.2 Convergence tests for series of constant There are always two important questions to ask about a series. 1. Does the series converge? 2. If it converges, what is its sum?

$ 12.2 Convergence tests for series of constantI.PositiveSeries1. Definition8Zu, is called a positive seriesif u, ≥0.n=1For a positive series,iS,j is increasing obviously2. Bounded sum test8ZtApositive seriesu, converges if and only if its partialn=1sums are bounded above.NoteThe convergence of apositiveseriesdoes notchangeby grouping inanyway
§12.2 Convergence tests for series of constant I. Positive Series is called a positive seriesif 0. 1 = n n un u 1. Definition For a positive series,{ }is increasing obviously. Sn 2. Bounded sum test sums are bounded above. A positive series convergesif and only if its partial 1 n= un Note The convergence of a positive series does not change by grouping in any way

$ 12.2 Convergence tests for series of constantI.PositiveSeries3. Comparison testTh Suppose that O ≤u, ≤ y, for n ≥ N8080ZZi) Ifconverges, sodoesVun=1n=188ZZii)Ifdiverges, sodoes1u1nn=1n=1AnalysisA, =u +u, +...+un B, =v +v +...+yA, ≤B
§12.2 Convergence tests for series of constant I. Positive Series 3. Comparison test Th Suppose that 0 un vn for n N i)If converges, s odoes . 1 1 = = n n n vn u ii) If diverges, s odoes . 1 1 = = n n n n u v An = u1 + u2 ++ un n n B = v + v ++ v 1 2 An Bn Analysis

S 12.2 Convergence tests for series of constantLPositiveSeries88ZZ70E.g.1K22n-1+Kn=n=11Solution2"2"+k812元>soconvergesbecauseconverges2" +kn=112n-12n801ZZdivergesbecausediverges.SO:2n2n-1n=1
§12.2 Convergence tests for series of constant I. Positive Series E.g.1 ( 0) 2 1 1 + = k n k n 2 1 1 1 n= n − Solution n n k 2 1 2 1 + n 2n 1 2 1 1 − converges. 2 1 converges because 2 1 s o 1 1 = = + n n n n k diverges. 2 1 diverges because 2 1 1 s o 1 1 = n= n − n n

S 12.2 Convergence tests for series of constantI.PositiveSeries80181ZE.g.221Zn(n+ 1)In!/4n2+25n=1n=1=1111Analysisn+1/n(n+1)hn(n+1)11110n/4n2+252nV4n2+25111n!1.2.3..n2.2.2...22n
§12.2 Convergence tests for series of constant I. Positive Series E.g.2 4 25 1 1 2 n= n + ! 1 1 ( 1) n= n 1 1 n= n n + Analysis n n n 1 ( 1) 1 + 1 1 ( 1) 1 + n n + n n 2n 1 4 25 1 2 + n 10n 1 4 25 1 2 + n n = 1 2 3 1 ! 1 2 2 2 2 1 n 2 1

S 12.2 Convergence tests for series of constantI.Positive Series8E.g.3>The series2n=1is called a p-series.Show that it converges if p > 1and diverges if p ≤1.Analysis-2If p=1,diverges1If p1
§12.2 Convergence tests for series of constant E.g.3 and diverges if 1. is called a -series.Show that it convergesif 1 1 3 1 2 1 1 1 The series 1 = + + + + + = p p p n n p p p n p I. Positive Series Analysis = = = = 1 1 1 1 If 1, n n p n n p n n p p 1 1 If 1, diverges diverges If p 1

$ 12.2 Convergence tests for series of constantI.PositiveSeries8E.g.3SThe seriesn=1is called a p-series.Show that it converges if p > 1and diverges if p ≤ 1.Analysis80Zu281b9Dn=18Zn=18011-Z<w.WUconverges4200n=1n=l
§12.2 Convergence tests for series of constant E.g.3 and diverges if 1. is called a -series.Show that it convergesif 1 1 3 1 2 1 1 1 The series 1 = + + + + + = p p p n n p p p n p I. Positive Series Analysis 15 1 9 1 8 1 7 1 6 1 5 1 4 1 3 1 2 1 1 1 = + + + + + + + + ++ + = p p p p p p p p p n un 15 1 9 1 8 1 7 1 6 1 5 1 4 1 3 1 2 1 1 1 + + + + + + + + + = + + = p p p p p p p p p n n v 8 1 4 1 2 1 1 1 1 1 1 = + − + − + − + = p p p n wn , n wn v converges. 1 n= un

$ 12.2 Convergence tests for series of constantLPositiveSeries8E.g.3>The seriesnn=]is called a p-series.Show that it converges if p > 1and diverges if p ≤ 1.801ZZForinstance,n(n +1)n=lVn
§12.2 Convergence tests for series of constant E.g.3 and diverges if 1. is called a -series.Show that it convergesif 1 1 3 1 2 1 1 1 The series 1 = + + + + + = p p p n n p p p n p I. Positive Series 1 1 n= n ( 1) 1 1 n= n n + For instance

$ 12.2 Convergence tests for series of constantL.PositiveSeries3. The Limit Comparison Test儿n =l(0 0, and limn->0 Vn1Eu, and Ev,converge or diverge together.Then808ZZIf 1 = 0andWnconverges, then2converges.7n=1n=188Z2Ifl = oanddiverges, thendiverges.unn=1n=1
§12.2 Convergence tests for series of constant 3. The Limit Comparison Test I. Positive Series and lim = (0 ), → l l v u n n n If and l = 0 converges, then converges. n=1 n v n=1 un If and l = diverges, then diverges. n=1 n v n=1 un Suppose that 0, 0, n n u v Then and converge or diverge together. un n v
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