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Unformatted text preview: 812005The Ratio Test and the Root TestIt is a fact that an increasing sequence of real numbers that is bounded above must converge.MThe picture shows an increasing sequence that is bounded above by some numberM. It seems reasonablethat the terms of the sequence will have to start piling up below the line, and at the place where they pileup the sequence will converge. A careful proof of this fact uses a deep property of the real numbers calledtheLeast Upper Bound Axiom.Supposesummationdisplayk=1akis a series with positive terms. The partial sumsa1, a1+a2, a1+a2+a3, . . .obviously form an increasing sequence.If the partial sums are bounded above,a1, a1+a2, a1+a2+a3, . . .Mfor someM, then the fact above implies that the series converges.Now letsummationdisplayk=1akbe a series with positive terms. LetL= limkak+1ak.TheRatio Testsays:IfL <1, the series converges.IfL >1, the series diverges.IfL= 1, the test fails.The reason the test works is that, in the limit, the series looks like a geometric series with ratioL. Lookat the caseL <1 by way of example. Choose a positive numbersor=L+ <1. Fornsufficiently large,an+1an,an+2an+1, . . . < r.These inequalities givean+1< ran,an+2< ran+1< r2an,an+3< ran+2< r3an,...1Adding the inequalities yieldsan+1+an+2+an+3+< ran+r2an+r3an+.The right side is a convergent geometric series. The inequality shows that its sum is an upper bound forthe partial sums of the series on the left. By the fact I stated at the start, the series on the left converges.Hence, the original seriessummationdisplayk=1akconverges, since its just(a1+a2++an) +an+1+an+2+an+3+.This is a finite number (a1+a2++an) plus the seriesan+1+an+2+an+3+, which I knowconverges.A similar argument works ifL >1.When do you use the Ratio Test? Ratios are fractions, and they tend to simplify nicelyif the top andbottom contain products or powers. For example, if thenthterm of the series containsfactorials, you oughtto give the Ratio Test serious consideration....
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 Spring '07
 buchanan
 Real Numbers

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