cours numerical series

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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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Calculus

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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)(a n ) n∈N , we can define the sequence(s n ) n∈N of partial sums as n ∀n∈N s n = X a k k=0 Furthermore, given a sequence of numbers(s n ) n∈N , we can define the sequence(a n ) n∈N as a 0 =s 0  ∀n∈N a n+1 =s n+1 −s n It is interesting to associate a sequence(a n ) n∈N and a sequence(s n ) n∈N , because properties of numerical series can be considered as properties on numerical sequences.

1 Definitions and examples Definition 1.Let(a n ) n∈N a sequence of numbers (real or complex). If the sequence(s n ) n∈N converges tos, we say that the series of(a n ) n∈N converges and we write ∞ X a n =s n=0 We call the numbersthe sum of the series. If the sequence(s n ) n∈N diverges, we say that the series diverges. ∞ Example .The simplest of all series is perhaps the geometric series X x n . We know that n=0 ∀x6= 1 N 1−x N+1 X x n = 1−x n=0 ∞ x n 1 •if|x|<1, then(x n ) n∈N →0and so X 1 = −x . n=0 ∞ •if|x| ≥1then(x n ) n∈N diverges and so X x n diverges. n=0

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∞ Theorem 1 (Cauchy criterion).The series X a n converges if, and only if, n=0 q ∀ǫ >0∃N∈N∀p∈N∀q∈N N≤p≤q⇒ | X a k |< ǫ k=p Proof .This is just the Cauchy criterion applied to the sequence(s n ) n∈N . ∞ Corollary .If the series X a k converges, the sequence(a n ) n∈N converges 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 ies X lndiverges. ser ∞ ( k k + 1  ) k=1 Theorem 2.of non negative terms is convergent if, and only if , the sequenceA series (s n ) n∈N is bounded. Proof .The sequence(s n ) n∈N will 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 integerN 0 , we have∀n≥N 0 |a n | ≤c n , and if the series X c n is convergent, then the series X a n is convergent too. Proof .Let X c n  Letbe a convergent series.ǫ >0. By the Cauchy criterion, there exists N 1 such that p ∀p≥q≥N 1  X c k < ǫ k=q PutN=max(N 0  N 1 ). Then we have p p p ∀p≥q≥N| X a k | ≤ X |a k | ≤ X c k < ǫ k=q k=q k=q The Cauchy criterion gives the result. Theorem 4.If, for some integerN 0 , we have∀n≥N 0  a n ≥d n ≥0, and if the series X d n diverges, then the series X a n diverges.

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Exercise1.Prove this theorem. Definition 2.A series X a k is said to be absolutely convergent if the series X |a k |con- verges. Exercise2.Proove that a absolutely convergent series is convergent. Corollary (Comparison tests for positive series).Let X a n and X c n two positive series. a n •If the series X c n is convergent, and if there existsαsuch thatlim =α, then the n→∞ c n series X a n is convergent too. •If the series X c n is divergent, and if there exists nonzeroαsuch that n li → m ∞ a c n n =α, then the series X a n is 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| a n c n −α|< ǫ Thus we have ∀n≥N a n < c n (α+ǫ) which proves the result by the previous theorem. Theorem 5 (Cauchy condensation test).Let(a n ) n∈N a decreasing nonnegative sequence. ∞ Then the series X a n converges if, and only if, the series X 2 k a 2 k converges k=0 Proof .positive termes, it is sufficient to prove that the sequencesSince it is a series of n K s n = X a k andt K = X 2 k a 2 k have the same comportment at the infinity. k=0k=0 Ifn≤2 K+1 1, then − s n ≤a 0 +a 1 + (a 2 +a 3 ) ++ (a 2 K ++a 2 K+1 −1 ) s n ≤a 0 +a 1 + 2a 2 ++ 2 K a 2 K s n ≤a 0 +t K

Ifn >2 K , then s n ≥a 1 +a 2 + (a 3 +a 4 ) ++ (a 2 K−1 +1 ++a 2 K ) s n ≥ 1 2 a 1 +a 2 + 2a 4 ++ +2 K−1 a 2 K = 1 2 t K Thus the sequences(s n )and(t K )are either both bounded or both unbounded. 4

Example .Here are some classical examples. 1. The series X n 1 s converges ifs >1and diverges ifs≤1. Indeed, we have by the n previous therorem, that this series has the same behaviour as the series X ∞ 2 k (2 k 1 ) s = k= X ∞ 0 2 k(1−s) k=0 But this is a geometric series withx= 2 1−s , and so this series is convergent if, and only if,s >1. ∞ 1 s if, and only ifs >1. Again, we loo 2. The series X k at the series n=2 n(log(n)) s converge X ∞ 2 k 2()l 1 go s X ∞ k 1 s = k=1 2 k (log(2 k )) s k=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 123 + 233 + 444 + 55 + 6  i.e.a n = n(n 2 +1 n ) − (n)2+ 1 . Letc n = n 1 2 . We have 2− 1 a n n c n +1( = 1 n )(1 + n 2 ) →2 Thus the series X a n and X c n have the same comportment. By the first item, we have that X a n converges. 4. We can look at series with more complicated terms. As an example, we can consider X n 2 1 1+ . Leta n = n 1 2 +1 andc n = n . Since c n , 1a n →1we have by the first item n that the series X a n diverges. We see on these two last examples that the series X n 1 s are 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 series X a n where : n=1 1.a n =n+ 1−n 2.a n n+1−n = n 3.a n = 3 n 3 + 1−n 4.a n = 4 n 4 + 1− 4 n 4 −1 1 5.a n = nlog(n)log(log(n)) 6.a n = n 2 1 +x 2 , wherexis a real number. 7.a n = sin( n 1 ) 8.a n = 1 n sin( n 1 ) 9.a n = sin( 1 n ) 2 10.a n = n+ n 1 p −n , wherepis a real number. Exercise4(Cauchy-Schwarz inequality).Prove that if a series X a n converges, then the series X a n n converges too. Indication : Show the Cauchy-Schwarz inequality n n n ∀n∈N X a k b k ≤ u a 2 k=1 t k= X 1 k t u k= X 1 b k 2

3 Integral Test Theorem 6.If X a n is a series of positive terms andf: [a∞)→Ris a positive monotoni- cally decreasing function andf(n) =a n for alln > a, then the improper integral R a ∞ f(x)dx ∞ and the series X a n converge or diverge together. n=1 Proof .Since∀n∈N a n >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) = R n K f(t)dt n J(n) = X a k k=K Sincefis positive and monotically decreasing, consider[k k+ 1], withk≥K, we obtain k+1 a k ≥ Z k f(t)dt≥a k+1 which implies by addition 1 X a k k= X n K a k ≥ Z n K +1 f(t)dt≥ k=K n 1+ +

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 > K  I J ( ( n n) ) ≥ ≥ I J ( ( n n) +) − R a K n n+1 f(t)dt Set the integerK≥a. Letǫ hypothesis, there exists Bya positive real number. KN > such that∀t > N0≤f(t)<1, and so∀n > N0≤ R n n+1 f(t)dt <1 since. Besides, the series diverges, we havelim n→∞ J(n) =∞and so there existsM > Nsuch that ∀n > M J(n)> a k ǫ +1 , and so ∀ǫ >0∃M∀n > M  I J( (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 ∞ ∞ X a f(t k= k K ≥ Z K )dt≥ k= X ∞ K+1 a k = Example .1. Leta n og(n( 1 (logl)gon)) 2  then define the function. We nl f [3 : ∞[→R x֌ xlog(xl)gol(go 1 (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 Z 3 ∞ f(t)dt= Z log ∞ (log(3)) t 1 2 dt <∞ 2. Leta n = n e n . We then look at the functionf:x֌xe −x , which is decreasing and positive. Besides, its integral converges : Z 0 ∞ f(t)dt= [−te −t ] 0 ∞ + Z ∞ e −t dt= e 1 0 Hence the series converges.

4 Ratio test Here is theorems which give convergent result by looking at the ratio sequence( a n a + n 1 ) n . Theorem 7.The series X a n is 1. convergent ifli n m → s ∞ up| a n a +1 |<1 n 2. divergent iflim inf| a n+1 |>1 n→∞ a n Proof .1 n+1 |<1, then there . Ifli n m → s ∞ up| a a n existβ <1such that 1 < β ∃N∀n > N| a a n n+ | which easily implies that ∀n > N|a n |< β n ( |a β N N | ) and the right-hand term is a convergent geometric series. Hence, by the comparison tests, we have the result. 2. Iflim inf| a n+1 | ≥1there existsn 0 such that∀n > n 0 | a a n+ n 1 |>1, then the sequence n→∞ a n (a n ) n  Hence can’t converge to zero.the result, by contradiction with the Cauchy criterion. Remark .Iflim inf| a n a +1 | ≤1≤li n m → s ∞ up| a a n n+1 |, we cannot give positive or negative result. n→∞ n In fact, the series X n 1 and X n 1 2 both verify these conditions and do not have the same convergence properties.

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Theorem 8.Consider the series X a n of positive terms and suppose that α1 a n a + n 1 = 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 X ng(ol 1 n) and X nlo 1 g(n) 2 verify 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(|a n | 1n ) n . This will be very useful for studying power series ( See next chapter ). Theorem 9.Given a numerical series X a n , setα= limsup|a n | 1n . Then 1. Ifα <1, then the series converges. 2. Ifα >1, then the series diverges. t fails. It suffices to consider again the ser s 1 and Remark .Ifα= 1, then the tes ie X n 1 X n 2 . Proof .Ifα <1, then there existsNsuch that∀n > N|a n | 1n ≤β <1, and so|a n | ≤β 1N and the series converges by comparison theorem. Ifα >1then there exists a subsequence such that, |a n k | 1n k →α >1, and so the general term doesn’t converge to zero, and the Cauchy criterion fails. .1. Consider n Example X n+ 1 x n . We have a n+1 (n+ 1) 2 a n = 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|a n | →1.

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x n 2. Consider X n!  have that. We

a n+1 x →0 = a n n+ 1 and so the series converges for allx. This is known as the exponential series. 3. Consider the series of general terma n = ( n n−1) 1n  we can look at the term. Here a 1nn = n n−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 .Leta 2n = 1 2 n anda 2n+1 = 3 1 n  have that. We ( inflim l a n a n +1 = 0 imsup a n+1 =∞ a n so the ratio tests fails. However, we easily have that limsupa 1n n 1 = 2

which prooves convergence. Definition 3.series is a series of the formA power X a n z n (z∈CorR) Setα s == lim 1 .R ies.is called the up|a n | 1n andR α radius of convergence of the power ser n→∞ [See next result] Theorem 10.The series X a n z n converges if|z|< Rand diverges if|z|> R. Proof .It suffices to consider|a n z n | 1n =|a n | 1n |z|and to use the root test. Example .1. X n n z n has a radius of convergenceR= 0. 2R=∞for z n . the series X n! . 3.R= 1for the series X z n

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6 More refined tests Theorem 11 (Kummer’s Test).Let X a n be a positive series and{p n }be positive constants such that n li → m ∞  p n a n a + n 1 −p n+1  =L exists. Then, 1. if0< L≤+∞, X a n converges. 2. if−∞ ≤L <0and X p 1 diverges, X a n diverges. n Proof .1. LetL >0. Let0< r < L. Then∃N such that f or n≥N p n a n a + n 1 −p n+1 > r p N a N −p N+1 a N+1 > r a N+1 p N+1 a N+1 −p N+2 a N+2 > r a N+2 p N+m−1 a N+m−1 −p N+m a N+m > r a N+m adding p N a N −p N+m a N+m > r(s N+m −s N ) s N+m < s N + 1 r p N a N Let’s denotes N + 1 r p N a N as C. Then{s n }is bounded by C, and X a n converges. 2. IfL <0,∃Nsuch that∀n≥N, p n+1 a n+1 > p n a n   ≥p N a N =C ∀n≥N a≥ C n p n since P 1 diverges, so does X a n . p n Corollary (Raabes Test).Let X a n be a positive series and L= lim ∞  n  a a n n+1 −1  n→ exists. •If L>1 X a n converges •If L<1 X a n diverges •If L=1 the test fails

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