2.4.2 A Theorem About Absolutely Convergent Series

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1 0 Versio of August 27, 200 CHAPTER 2. INFINITE SERIES Add these two series: = 3 l 2. (2.20) 2 Sice the reciprocal of each iteger occurs exactly oce i the last series, we would be tempted to rearrage the series to obtai = l 2, (2.2) 6 which is idetical to the origial series. There is a obvious cotradictio here! I order to obtai the rearragemet (2.2), we have to go further ad further out i the series (2.20), which apparetly is ot permissible A Theorem About Absolutely Coverget Series Not oly ca absolutely coverget series be rearraged without chagig their value, but they ca be multiplied together term by term: If two series S = T = u i, i= i= v i (2.22a) (2.22b) are both absolutely coverget, the series P = u i v j (2.23) i= j= formed from the product of their terms writte i ay order, is absolutely coverget, ad has a value equal to the product of of the idividual series, 2.5 Covergece Tests P = ST. (2.24) The followig tests ca determie whether a give series is absolutely coverget or ot Compariso test If b > 0 for all ad b is coverget, ad if a b for all, the a is absolutely coverget. (2.25a)

2 2.5. CONVERGENCE TESTS Versio of August 27, 200 Also, if a b > 0 for all, ad b diverges, the Root test a is ot absolutely coverget. (2.25b) The series a coverges absolutely if from a certai term oward a q <, (2.26) where q 0 is idepedet of. Proof: If the iequality holds, a q. But q coverges for q <, it beig the geometric series, so by 2.5., a coverges Ratio test The series a coverges absolutely if from a certai term oward a q <, (2.27) where q 0 is idepedet of. Proof: Without loss of geerality, we may assume the iequality holds for all ; otherwise, we reumber the {a } sequece so that labels the first term for which the iequality (2.27) holds. The a = a a a 2 a 2 q. (2.28) a a a 2 a 3 Covergece is agai assured by compariso with the geometric series. (Whether these tests are satisfied by the first few terms of a series is immaterial, sice a fiite umber of terms of a ifiite seris has o effect o the covergece.) Example Whe does q coverge? If we use the root test, we examie a = q lim = q, lim a (2.29a) while if we use the ratio test, we look at lim + a = q lim = q. (2.29b) I either case, we see that the series is absolutely coverget if q <, ad diverget otherwise. Because l = l, which teds to zero as,.

3 2 Versio of August 27, 200 CHAPTER 2. INFINITE SERIES The followig are refiemets of the ratio test, which fails (that is, fails to reveal whether the tested series coverges or ot) whe lim a =. (2.30) For example, this idetermiate limit results for the case a = /, which yields a diverget series, but also for a = /( l 2 ), which correspods to a coverget sum (see Sec ) Kummer s test Choose a sequece of positive costats b. If a b b + C > 0, (2.3) for all N, where N ad C are fixed umbers, the O the other had, if ad the a coverges absolutely. (2.32) b a b + 0, (2.33) b diverges, (2.34) a diverges. (2.35) Proof: If the iequality (2.3) holds, take l N, so that So we have the iequality l=n+ C a l+ b l a l b l+ a l+. (2.36) a l b N a N C Hece, the th partial sum, for > N, is s = i= b a C b N a N C. (2.37) N a i a i + b N a N C. (2.38) i=

4 2.5. CONVERGENCE TESTS 3 Versio of August 27, 200 The right-had side of this iequality is a costat, idepedet of. Therefore, the positive sequece of icreasig terms {s } is bouded above, ad cosequetly possesses a limit. The series is absolutely coverget. If the iequality (2.33) holds, so sice b diverges, so does a Raabe s test a a N b N b, > N, (2.39) Raabe s criterio for absolute covergece is ( ) a K >, (2.40) for all N, where N ad K are fixed. Ad if ( ) a, (2.4) the a diverges. (2.42) Proof: I Kummer s test put b = Gauss test If a = + h + B() 2, (2.43) where h is a costat ad the fuctio B() is bouded as, the a coverges for h > ad diverges for h. Proof: For h we ca use Raabe s test: ( h lim + B() ) 2 = h. (2.44) For h =, Raabe s test is idetermiate. I that case use Kummer s test with b = l : for large, ( l + h + B() ) 2 ( + ) l( + ) ( l + h + B() ) ( 2 ( + ) l + ) ( h + B() 2 ) l l (h ) l < 0, if h. (2.45)

5 4 Versio of August 27, 200 CHAPTER 2. INFINITE SERIES f Figure 2.: Bouds o a mootoe series provided by a itegral. Because =2 l (see homework), the series a diverges Itegral test diverges (2.46) If f(x) is a cotiuous, mootoically decreasig real fuctio of x such that the a coverges if ad diverges otherwise. Proof: It is geometrically obvious that dx f(x) < f() < f() = a, (2.47) dx f(x) <, (2.48) dx f(x) + f(), (2.49) for this follows merely from the geometrical meaig of the itegral as the area uder the curve of the fuctio. See Fig Examples The Riema zeta fuctio is defied by the series = ζ(α). (2.50) α We ca test for covergece usig Gauss test, by examiig ( ) α + + α for large. (2.5) Thus the series coverges if α >, ad diverges if α.

6 2.5. CONVERGENCE TESTS 5 Versio of August 27, 200 Cosider the series (l ) α. (2.52) Let s use Raabe s test: ( ) α ( ) α l( + ) l + l( + /) = + α l l l. (2.53) Because ( α ) = α 0 l l as, (2.54) we coclude that the series is diverget. To test for covergece of let us use the itegral test: =2 (l ) α, (2.55) 2 dx x(l x) α = = = l 2 d(l x) (l x) α α (l x) α x=2, α, l(l x) x=2, α = { fiite α >, α. (2.56) Thus the series coverges if α > ad diverges for other real α.

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