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Calculus IIAxelle ZieglerSylvain ErvedozaXavier GendreRodolphe Richardfrom courses notes by S. Kesavan16Part INumerical seriesIntroductionGiven a sequence of numbers (real or complex) (a ) , we can define the sequence (s )n n∈N n n∈Nof partial sums asnX∀n∈N,s = an kk=0Furthermore, given a sequence of numbers (s ) , we can define the sequence (a )n n∈N n n∈Nas a =s0 0∀n∈N,a = s −sn+1 n+1 nItisinterestingtoassociateasequence (a ) andasequence (s ) ,becausepropertiesn n∈N n n∈Nof numerical series can be considered as properties on numerical sequences.1 Definitions and examplesDefinition 1. Let (a ) a sequence of numbers (real or complex).n n∈NIf the sequence (s ) converges to s, we say that the series of (a ) converges and wen n∈N n n∈Nwrite ∞Xa = snn=0We call the number s the sum of the series.If the sequence (s ) diverges, we say that the series diverges.n n∈N∞XnExample . The simplest of all series is perhaps the geometric series x . We know thatn=0N N+1X 1−xn∀x = 1, x =1−xn=0∞X 1n n• if|x| < 1, then (x ) → 0 and so x = .n∈N1−xn=0∞Xnn• if|x|≥ 1 then (x ) diverges and so x diverges.n∈Nn=02∞XTheorem 1 (Cauchy criterion). The series a converges if, and only if,nn=0qX∀ǫ > 0,∃N ∈N,∀p∈N,∀q∈N, N ≤p≤q⇒| a | <ǫkk=pProof . This is just the Cauchy criterion applied to the sequence (s ) .n n∈N∞XCorollary . If the series a converges, the sequence (a ) converges to zero.k n n∈Nk=0Proof . The Cauchy criterion gives the result in taking q ...
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Axelle Ziegler Sylvain Ervedoza Xavier Gendre Rodolphe Richard courses notes by S. Kesavan
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Part I Numerical series Introduction Given a sequence of numbers (real or complex)(an)n∈N, we can define the sequence(sn)n∈N of partial sums as n ∀n∈N sn=Xak k=0 Furthermore, given a sequence of numbers(sn)n∈N, we can define the sequence(an)n∈N as a0=s0 ∀n∈N an+1=sn+1−sn It is interesting to associate a sequence(an)n∈Nand a sequence(sn)n∈N, because properties of numerical series can be considered as properties on numerical sequences.
1 Definitions and examples Definition 1.Let(an)n∈Na sequence of numbers (real or complex). If the sequence(sn)n∈Nconverges tos, we say that the series of(an)n∈Nconverges and we write ∞ Xan=s n=0 We call the numbersthe sum of the series. If the sequence(sn)n∈Ndiverges, we say that the series diverges. ∞ Example .The simplest of all series is perhaps the geometric seriesXxn. We know that n=0 ∀x6= 1N1−xN+1 Xxn=1−x n=0 ∞xn1 •if|x|<1, then(xn)n∈N→0and soX1=−x. n=0 ∞ •if|x| ≥1then(xn)n∈Ndiverges and soXxndiverges. n=0
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∞ Theorem 1 (Cauchy criterion).The seriesXanconverges if, and only if, n=0 q ∀ǫ >0∃N∈N∀p∈N∀q∈N N≤p≤q⇒ |Xak|< ǫ k=p Proof .This is just the Cauchy criterion applied to the sequence(sn)n∈N. ∞ Corollary .If the seriesXakconverges, the sequence(an)n∈Nconverges to zero. k=0 Proof .The Cauchy criterion gives the result in takingq=p. Remark . a counterexample, we can easily see that theThis condition is not sufficient. As iesXlndiverges. ser∞(kk+ 1 ) k=1 Theorem 2.of non negative terms is convergent if, and only if , the sequenceA series (sn)n∈Nis bounded. Proof .The sequence(sn)n∈Nwill be a monotonic sequence. Remark .divergence) of a numerical series will not be affected byThe convergence (or the the first terms. These terms only affect the sum of the series.
2 Comparison tests Theorem 3.If, for some integerN0, we have∀n≥N0|an| ≤cn, and if the seriesXcnis convergent, then the seriesXanis convergent too. Proof .LetXcn Letbe a convergent series.ǫ >0. By the Cauchy criterion, there exists N1such that p ∀p≥q≥N1Xck< ǫ k=q PutN=max(N0 N1). Then we have p p p ∀p≥q≥N|Xak| ≤X|ak| ≤Xck< ǫ k=q k=q k=q The Cauchy criterion gives the result. Theorem 4.If, for some integerN0, we have∀n≥N0 an≥dn≥0, and if the seriesXdn diverges, then the seriesXandiverges.
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Exercise1.Prove this theorem. Definition 2.A seriesXakis said to be absolutely convergent if the seriesX|ak|con-verges. Exercise2.Proove that a absolutely convergent series is convergent. Corollary (Comparison tests for positive series).LetXanandXcntwo positive series. an •If the seriesXcnis convergent, and if there existsαsuch thatlim =α, then the n→∞cn seriesXanis convergent too. •If the seriesXcnis divergent, and if there exists nonzeroαsuch thatnli→m∞acnn=α, then the seriesXanis divergent too. Proof .We will prove just the first part of the corollary, the proof of the second item is similar. Letǫ >0. By hypothesis, there existsNsuch that ∀n≥N|an cn−α|< ǫ Thus we have ∀n≥N an< cn(α+ǫ) which proves the result by the previous theorem. Theorem 5 (Cauchy condensation test).Let(an)n∈Na decreasing nonnegative sequence. ∞ Then the seriesXanconverges if, and only if, the seriesX2ka2kconverges k=0 Proof .positive termes, it is sufficient to prove that the sequencesSince it is a series of n K sn=XakandtK=X2ka2khave the same comportment at the infinity. k=0k=0 Ifn≤2K+11, then − sn≤a0+a1+ (a2+a3) ++ (a2K++a2K+1−1) sn≤a0+a1+ 2a2++ 2Ka2K sn≤a0+tK
Ifn >2K, then sn≥a1+a2+ (a3+a4) ++ (a2K−1+1++a2K) sn≥12a1+a2+ 2a4++ +2K−1a2K=12tK Thus the sequences(sn)and(tK)are either both bounded or both unbounded. 4
Example .Here are some classical examples. 1. The seriesXn1sconverges ifs >1and diverges ifs≤1. Indeed, we have by the n previous therorem, that this series has the same behaviour as the series X∞2k(2k1)s=k=X∞02k(1−s) k=0 But this is a geometric series withx= 21−s, and so this series is convergent if, and only if,s >1. ∞1s if, and only ifs >1. Again, we loo 2. The seriesXk at the series n=2n(log(n))sconverge X∞2k2()l1gosX∞k1s = k=12k(log(2k))sk=1 and we deduce the result by the previous item. 3. We look at the convergence of the series defined by 1 3 5 7 123+233+444+55+6 i.e.an=n(n2+1n)−(n)2+1. Letcn=n12. We have 2−1 ann cn+1(=1n)(1 +n2)→2 Thus the seriesXanandXcnhave the same comportment. By the first item, we have thatXanconverges. 4. We can look at series with more complicated terms. As an example, we can consider Xn211+. Letan=n12+1andcn=n. Sincecn, 1an→1we have by the first item n that the seriesXandiverges. We see on these two last examples that the seriesXn1sare very often used to prove n convergence or divergence of numerical series. In fact, the problem is that we can use comparison theorem only if we know the behaviour of many series, for instance these series, which convergence is easy to prove.
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∞ Exercise3.Test for convergence of the nunerical seriesXanwhere : n=1 1.an=n+ 1−n 2.an n+1−n = n 3.an=3n3+ 1−n 4.an=4n4+ 1−4n4−1 1 5.an=nlog(n)log(log(n)) 6.an=n21+x2, wherexis a real number. 7.an= sin(n1) 8.an=1nsin(n1) 9.an= sin(1n)2 10.an=n+n1p−n, wherepis a real number. Exercise4(Cauchy-Schwarz inequality).Prove that if a seriesXanconverges, then the seriesXannconverges too. Indication : Show the Cauchy-Schwarz inequality n n n ∀n∈NXakbk≤ua2 k=1tk=X1ktuk=X1bk2
3 Integral Test Theorem 6.IfXanis a series of positive terms andf: [a∞)→Ris a positive monotoni-cally decreasing function andf(n) =anfor alln > a, then the improper integralRa∞f(x)dx ∞ and the seriesXanconverge or diverge together. n=1 Proof .Since∀n∈N an>0andf >0the improper integral (which is defined as, both the limit of the integral fromatokwith k goes to the infinity) and the limit of the series, exist if we accept the infinity value. So we only have to prove that the integral is bounded if,
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and only if, the partial sums of the series are uniformly bounded. LetK > aa fixed integer. We define forn > K I(n) =RnKf(t)dt n J(n) =Xak k=K Sincefis positive and monotically decreasing, consider[k k+ 1], withk≥K, we obtain k+1 ak≥Zkf(t)dt≥ak+1 which implies by addition 1 Xak k=XnKak≥ZnK+1f(t)dt≥k=Kn1++
Hence the result. Remark .1. We have proved a stronger result if the limit offat the infinity is zero, which is the interesting case of this theorem because of the Cauchy criterion, and if both the integral and the series diverge. Namely, ∀K≥a∀n > KIJ((nn))≥≥IJ((nn)+)−RaKnn+1f(t)dt Set the integerK≥a. Letǫ hypothesis, there exists Bya positive real number. KN > such that∀t > N0≤f(t)<1, and so∀n > N0≤Rnn+1f(t)dt <1 since. Besides, the series diverges, we havelimn→∞J(n) =∞and so there existsM > Nsuch that ∀n > M J(n)>akǫ+1, and so ∀ǫ >0∃M∀n > MIJ((n))−1< ǫ n We say thatI(n)andJ(n)are equivalent in the infinity. 2. In fact, the termsI(n)andJ(n)depend on the numberK, and we will note these termes respectivelyI(n K)andJ(n K). If we look at the previous inequalities, and if the series (or the integral) is convergent, then both terms converge together whenn goes to the infinity, and we obtain ∞∞ Xa f(t k=kK≥ZK)dt≥k=X∞K+1ak = Example .1. Letanog(n(1(logl)gon))2 then define the function. We nl f[3:∞[→R xxlog(xl)gol(go1(x))2 7
This function is positive decreasing, and we can thus apply the theorem and deduce that the series converges. In fact, we have Z3∞f(t)dt=Zlog∞(log(3))t12dt <∞ 2. Letan=nen. We then look at the functionf:xxe−x, which is decreasing and positive. Besides, its integral converges : Z0∞f(t)dt= [−te−t]0∞+Z∞e−tdt=e1 0 Hence the series converges.
4 Ratio test Here is theorems which give convergent result by looking at the ratio sequence(ana+n1)n. Theorem 7.The seriesXanis 1. convergent iflinm→s∞up|ana+1|<1 n 2. divergent iflim inf|an+1|>1 n→∞an Proof .1n+1|<1, then there . Iflinm→s∞up|aanexistβ <1such that 1< β ∃N∀n > N|aann+| which easily implies that ∀n > N|an|< βn(|aβNN|) and the right-hand term is a convergent geometric series. Hence, by the comparison tests, we have the result. 2. Iflim inf|an+1| ≥1there existsn0such that∀n > n0|aan+n1|>1, then the sequence n→∞an (an)n Hence can’t converge to zero.the result, by contradiction with the Cauchy criterion. Remark .Iflim inf|ana+1| ≤1≤linm→s∞up|aann+1|, we cannot give positive or negative result. n→∞n In fact, the seriesXn1andXn12 both verify these conditions and do not have the same convergence properties.
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Theorem 8.Consider the seriesXanof positive terms and suppose that α1 ana+n1= 1−+o(n) n 1. ifα <1, then the series diverges. 2. ifα >1, then the series converges. We will see the proof later [See Raabe’s test]. Remark .Ifα= 1 In, then we cannot conclude like in the previous case. fact, the series Xng(ol1n)andXnlo1g(n)2verify these properties and do not have the same convergence properties.
5 Root test Here we see theorems which give convergence criterions by looking at the sequence(|an|1n)n. This will be very useful for studying power series ( See next chapter ). Theorem 9.Given a numerical seriesXan, setα= limsup|an|1n. Then 1. Ifα <1, then the series converges. 2. Ifα >1, then the series diverges. t fails. It suffices to consider again the ser s1and Remark .Ifα= 1, then the tes ieXn 1 Xn2. Proof .Ifα <1, then there existsNsuch that∀n > N|an|1n≤β <1, and so|an| ≤β1N and the series converges by comparison theorem. Ifα >1then there exists a subsequence such that, |ank|1nk→α >1, and so the general term doesn’t converge to zero, and the Cauchy criterion fails. .1. Considern ExampleXn+ 1xn. We have an+1(n+ 1)2 an=n(n+ 2)x→x So, if|x|<1then the series converges, and if, |x|>1 If, the series diverges.x= 1, then the test fails. However, it diverges because|an| →1.
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xn 2. ConsiderXn! have that. We
an+1x→0 = ann+ 1 and so the series converges for allx. This is known as the exponential series. 3. Consider the series of general terman= (nn−1)1n we can look at the term. Here a1nn=nn−1→0and so the series converges. Remark .The root test is more powerful than the ratio test, but it is also more difficult to apply. Example .Leta2n=12nanda2n+1=31n have that. We (inflimlanan+1= 0 imsupan+1=∞ an so the ratio tests fails. However, we easily have that limsupa1nn1= 2
which prooves convergence. Definition 3.series is a series of the formA power Xanzn(z∈CorR) Setα s == lim1.R ies.is called the up|an|1nandRαradius of convergence of the power ser n→∞ [See next result] Theorem 10.The seriesXanznconverges if|z|< Rand diverges if|z|> R. Proof .It suffices to consider|anzn|1n=|an|1n|z|and to use the root test. Example .1.Xnnznhas a radius of convergenceR= 0. 2R=∞forzn . the seriesXn!. 3.R= 1for the seriesXzn
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6 More refined tests Theorem 11 (Kummer’s Test).LetXanbe a positive series and{pn}be positive constants such that nli→m∞pnana+n1−pn+1=L exists. Then, 1. if0< L≤+∞,Xanconverges. 2. if−∞ ≤L <0andXp1diverges,Xandiverges. n Proof .1. LetL >0. Let0< r < L. Then∃N such that f or n≥N pnana+n1−pn+1> r pNaN−pN+1aN+1> r aN+1 pN+1aN+1−pN+2aN+2> r aN+2 pN+m−1aN+m−1−pN+maN+m> r aN+m adding pNaN−pN+maN+m> r(sN+m−sN) sN+m< sN+1rpNaN Let’s denotesN+1rpNaNas C. Then{sn}is bounded by C, andXanconverges. 2. IfL <0,∃Nsuch that∀n≥N, pn+1an+1> pnan ≥pNaN=C ∀n≥N a≥C npn sinceP1diverges, so doesXan. pn Corollary (Raabes Test).LetXanbe a positive series and L= lim∞naann+1−1 n→ exists. •If L>1Xanconverges •If L<1Xandiverges •If L=1 the test fails
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