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Chapter 8Series IIExplicit SumsFor most convergent seriesthere is no simple formula forthe sumP∞n=1anin termsof standard mathematical ob-jects. Only in very lucky casescan we sum the series explic-itly, for instance geometric se-ries, telescoping series, variousseries found by contour inte-gration or by Fourier expan-sions. But these cases are souseful and so much fun that wemention them often.Usually we are doomed to failure if we seek a formula for the sum of a series.Nevertheless we can often tell whether the series converges or diverges withoutexplicitly finding the sum. To do this we shall establish a variety of convergencetests that allow us in many cases to work out from the formula for the termsanwhether the series converges or not.8.1 Series with positive termsSeries with positive terms are easier than general series since the partial sums(sn) form an increasing sequence and we have already seen that monotonicsequences are easier to cope with than general sequences.All our convergence tests are based on the most useful test - the comparisontest - which you have already proven.Sometimes the series of which we want to find the sum looks quite compli-cated. Often the best way to find a series to compare it with is to look at whichterms dominate in the original series.ExampleConsider the seriesP√n+2n3/2+1. We can rearrange thenthterm in thisseries as follows:√n+ 2n3/2+ 1=1 +2√nn+1√n.Asngets large then1√ngets small so the dominant term in the numerator isthe 1 and in the denominator is then. Thus a possible series to compare itLook Where You’reGoing!As with many of the resultsin this course, the ComparisonTest requires you to know inadvance whether you are try-ing to prove convergence or di-vergence. Otherwise you mayend up with a comparison thatis no use!with isP1n. Since this diverges, we want to show that our series is greater thansome multiple ofP1n:√n+ 2n3/2+ 1=1 +2√nn+1√n>1 +2√n2n>12n.hence by the comparision test,P√n+2n3/2+1diverges.73
74CHAPTER 8. SERIES IIExercise 1Use the Comparison Test to show thatP∞n=11npconverges ifp∈[2,∞) and diverges ifp∈(0,1]. [Hint - you already know the answer forp= 1 or 2.]Exercise 2Use this technique with the Comparison Test to determine whethereach of the following series converges or diverges. Make your reasoning clear.1.∞Xn=11n(n+ 1)(n+ 2)2.∞Xn=15n+ 4n7n−2nExercise 3Use the Comparison Test to determine whether each of thefollowing series converges or diverges.1.∞Xn=113√n2+ 12.∞Xn=113√n7+ 13.∞Xn=1√n+ 1−√n
8.2 Ratio TestThe previous tests operate by comparing two series. Choosing a GeometricSeries for such a comparison gives rise to yet another test which is simple andeasy but unsophisticated.The Missing CaseThe caseℓ= 1 is omittedfrom the statement of the Ra-tio Test. This is because thereexist both convergentanddi-vergent series that satisfy thiscondition.A.K.A.This test is also calledD’Alembert’s Ratio Test, afterthe French mathematicianJean Le Rond D’Alembert(1717 - 1783). He developedit in a 1768 publication inwhich he established theconvergence of the BinomialSeries by comparing it withthe Geometric Series.TheoremRatio TestSupposean>0 for alln≥1 andan+1an→ℓ. ThenPanconverges if 0≤ℓ <1and diverges ifℓ >1.Examples1. Consider the seriesP1n!. Lettingan=1n!we obtainan+1an=n!(n+1)!=1n+1→0. ThereforeP1n!converges.2. Consider the seriesPn22n. Lettingan=n22nwe obtainan+1an=(n+1)2n2·2n2n+1=121 +1n
2→12. ThereforePn22nconverges.Proof.(a) Convergence ifℓ <1. Sincean+1an→ℓ <1, there existsNsuch thatan+1an<ℓ+12<1 for alln > N. (There is nothing special about the numberℓ+12,we only need a number that is strictly greater thanℓand strictly less than 1.)Thenan+1<ℓ+ 12an<ℓ+ 122an−1< ... <ℓ+ 12n−N+1aN.