11.6 Absolute Convrg. (Ratio & Root Tests) & 11.7 Strategy for Testing Series

Transcription

1 11.6 Absolute Covrg. (Ratio & Root Tests) & 11.7 Strategy for Testig Series

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3 Give ay series Σ a, we ca cosider the correspodig series 1 a a a a whose terms are the absolute values of the terms of the origial series.... 3

4 A series Σa is called absolutely coverget if the series of absolute values Σ a is coverget. Notice that, if Σa is a series with positive terms, the a = a. So, i this case, absolute covergece is the same as covergece. 4

5 The series 1 1 ( 1) is absolutely coverget because the series 1 ( 1) is a coverget p-series (p = 2). 5

6 A series Σ a is called coditioally coverget if it is coverget but ot absolutely coverget. 6

7 We kow that the alteratig harmoic series is coverget. 1 1 ( 1) It is ot absolutely coverget because the correspodig series of absolute values is: 1 ( 1) This is the harmoic series (p-series with p = 1) ad is, therefore, diverget. 7

8 It is possible for a series to be coverget but ot absolutely coverget (see previous example). However, the coverse is true (as will see i the ext theorem). i.e, absolute covergece implies covergece. 8

9 If a series Σ a is absolutely coverget, the it is coverget. 9

10 Determie whether the series 1 si 4 si 4 si8 si is coverget or diverget. SOLUTIONS 10

11 a 1 I. If lim L 1, the a is absolutely Coverget a 1 a 1 II. If lim L 1, or L=, the a is Diverget a 1 III. If a 1 lim 1, the Ratio test is Icoclusive: a that is, o coclusio ca be draw about the covergece or divergece of a 1 11

12 Determie whether the series is absolutely coverget, coditioally coverget or diverget SOLUTIONS

13 I. If lim a L1, the a is absolutely coverget 1 II. If lim a L 1 or lim a the the series a 1 is diverget. III. If lim a 1 the the Root Test is icoclusive. 13

14 Determie whether the series is absolutely coverget, coditioally coverget or diverget

15 If L = 1 i the Ratio Test, do t try the Root Test sice L will be 1 also. If L = 1 i the Root Test, do t try the Ratio Test sice it will fail too. 15

16 Determie whether the series is absolutely coverget, coditioally coverget, or diverget SOLUTION ! SOLUTIONS

17 We ow have several ways of testig a series for covergece or divergece. The problem is to decide which test to use o which series. It is ot wise to apply a list of the tests i a specific order util oe fially works. Istead, as with itegratio, the mai strategy is to classify the series accordig to its form. 17

18 If the series is of the form 1 p it is a p-series. So it will be coverget if p > 1 diverget if p 1 18

19 If the series has the form it is a geometric series. So it will coverge if r < 1 diverge if r ar or ar Some prelimiary algebraic maipulatio may be required to brig the series ito this form. 19

20 If the series has a form that is similar to a p-series or a geometric series, the oe of the compariso tests should be cosidered. I particular, if a is a ratioal fuctio or a algebraic fuctio of (ivolvig roots of polyomials), the the series should be compared with a p-series. The compariso tests apply oly to series with positive terms. If Σ a has some egative terms, we ca apply the Compariso Test to Σ a ad test for absolute covergece. 20

21 If you ca see at a glace that lim a 0 the Test for Divergece should be used. 21

22 If the series is of the form Σ ( 1) -1 b or Σ ( 1) b the Alteratig Series Test is a obvious possibility. 22

23 Series that ivolve factorials or other products (icludig a costat raised to the th power) are ofte coveietly tested usig the Ratio Test. Bear i mid that a +1 /a 1 as for all p- series ad, therefore, all ratioal or algebraic fuctios of. Thus, the Ratio Test should ot be used for such series. 23

24 If a is of the form (b ), the Root Test may be useful. 24

25 If a = f(), where 1 f ( x) dx is easily evaluated, the Itegral Test is effective. This is valid assumig the hypotheses of this test are satisfied. 25

26 26 Test the series for covergece or divergece. SOLUTIONS SOLUTIONS si k k k

27 Test the series for covergece or divergece a where a = ad a = a SOLUTIONS (2 1) (3 1) 27

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