: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