Chapter 1 : Numerical Series.

Transcription

1 Chapter : Numerical Series. Defiitio ad first properties Defiitio.. Give a sequece of real or complex umbers a (a, we defie the sequece (s (a of partial sums by s (a a k. k The series associated to a is deoted by the symbol a, a or just a, ad is said coverget if the sequece of partial sums coverges to a limit S called the sum of the series. I this case, it will be usefull to ote (r (a the sequece of the remaiders associated to the coverget series a, defied by r (a S s (a k+ The series a is said diverget if the sequece (s (a diverges. Examples.2. Telescopig series : a sequece (a ad the telescopig series (a + a have the same behaviour. Geometric series : for a fixed real umber x ±, we have k0 x k x+ x a k. ( x coverges x <. Theorem.3. [Cauchy criterio] The series a coverges if ad oly if p ɛ > 0, N N, p q N, a k ɛ. Proof. Cauchy criterio o (s (a. Corollary.4. A ecessary coditio for a to coverge is the covergece of a to 0. Proof. Cosider p q i the previous proof. kq

2 Example.5. ( diverges. Remark.6. The coditio is ot sufficiet : we ll see later that diverges. Notatios.7. I will ofte ote ( 0 istead of ( N N such that N. Theorem.8. Suppose 0, a 0. We have a coverges (s (a bouded. Proof. N such that (s (a N is a icreasig sequece. 2 Compariso tests Notatios 2.. [Ladau otatios] Let (a ad (b be two sequeces. (Big O We ote a O(b if K R, 0, a K b. (little o We ote a o(b if ɛ > 0, 0, a ɛ b. (equivalece of sequeces We ote a b if a b o(a. Exercise 2.2. Check that a b o(a a b o(b. Remark 2.3. Suppose 0, b > 0. The we have a O(b a bouded, a o(b a 0, a b a. b b b Be careful with the implicatio (a α, b β a + b α + β, it s false if 0, α + β 0 : it would mea that 0, a + b 0, which is obviously ot ecessarily true. Cosider a /( +, b /( + 2 ad α β / : i fact we have a + b /(( + ( + 2 / 2. I such cases, it s more safe to use equalities istead of equivaleces, for example with the o ad O otatios. Example 2.4. If a 0 we have a < 2 (itegratio by parts for big eough, ad we ca write hece thus +a + a t t 2 dt a l( + a a a +a +a +a dt t a l( + a, + a t dt t 2 dt t 2 a a 2 a 2, + a l( + a a + O(a 2 Theorem 2.5. Let (a ad (b be two sequeces with 0, b 0.. If a O(b, (i b coverges a coverges ad r (a O(r (b, (ii b diverges s (a O(s (b. 2. Same statemets with o. 3. If a b, a ad b have the same behaviour ad 2

3 (i r (a r (b i case of covergece, (ii s (a s (b i case of divergece. Proof. (Partial First, 3 directly follows from ad 2. Let s prove (i : ( +p ( K, 0, a Kb K, 0, p, a k K k+ +p k+ By Cauchy criterio, a coverges ad we ca make p to obtai the result. Let s suppose a o(b ad b diverget to prove 2(ii. We fix ɛ > 0 ad N such that for all N, a ɛb. The N s (a a k k }{{} costat K 0 +ɛ ( ( b k b k ɛ + kn k0 K k0 b. k But b diverges ad b 0 for big eough, so k0 b k ad there exists N N such that s (a 2ɛ s (b, which gives the expected result. Remark 2.6. We ca use the cotrapositio of these statemets, for example a diverges ad a O(b b diverges. b k. Examples 2.7. a ad we ll see i the 2 ext sectio that α coverges iff α >, so a coverges. + a sup [,+] f where f(x x, hece 0 a 2, so a coverges because a O( 3/2. Aother way to treat this kid of sequece where it appears somethig like f(a f(b, with f differetiable, is to write (here f a + (/ / f ( 2 ad we have a more precise result. By 2.4, we have ( ( l + coverget, ad we ca ote γ its sum (the Euler-Mascheroi costat. We ca rewrite it as ( {( } γ lim (l(k + l(k lim l( + k k k ad fially (cf. l( + l o( k l + γ + o( k k Note that it implies (/ / ad 2.5.3(ii. k k l, which is a direct cosequece of l( + 3

4 3 Itegral test Theorem 3.. Let f : [a, + [ R + be a cotiuous decreasig fuctio. The for all N a we have Proof. We write x lim x + f(tdt f( coverges. N N, f( + x Hece if lim x + f(tdt S, a a + f(k f(n + kn N f(tdt f(. ( f(tdt S ad f(k 0 so we ca use.8 to obtai the covergece of f(. Coversely, if N f( coverges to S, because F : x x f(tdt is a icreasig fuctio, a we just have to prove that F is bouded : but for all x a, there exists N max{x, N + } ad, usig (, F (x F (N+ x N which gives the result. N f(tdt F (N+ N N f(tdt F (N+ kn f(k F (N+S. Examples For α > 0, f α : x x α is cotiuous ad decreasig o x x α if α [, + [ ad F α (x t α dt α, which implies l x if α that α coverges iff α > (cf. for α 0, a 0, which is a ecessary coditio. Let s use 2.5 to fid a equivalet of r,α k+ α for α > : first we have to fid a iterestig equivalet for α, typically somethig telescopig to obtai a ice remaider. We rewrite the formula ( for f f α, which gives : ( + α ( α α ( + α α Multiplyig this lie by α, we remark ad usig k+ α we obtai (cf (i ( α ( + α α, ( k α (k + α ( + α α, r,α α α 4

5 2. [Bertrad series] Let a α l β. if α <, α ]α, [, ad α o(a. But α diverges so by 2.5, a diverges. if α >, α ], α[, ad a o( α. Hece, this time, a coverges. if α, β 0, O(a, so a diverges. [2, + [ R if α, β > 0, f is cotiuous, decreasig, ad x x l β x x 2 l x du f(tdt ul t l 2 u β F β(l x F β (l 2, which has a fiite limite iff β > (cf. first example. 4 Ratio tests Propositio 4.. Let (a be a sequece such that 0, a > 0. (i If α < such that 0, a+ a < α, the a coverges. (ii If 0, a+ a, the a diverges. Proof. For (i, N such that N, a + α a with α ]α, [. Thus N we have a α N a N which implies a O(α ad so the result. For (ii, a 0. Corollary 4.2. [De D Alembert rule] With the same hypothesis, a (i If lim + a <, the a coverges. a (ii If lim + a >, the a diverges. Remark 4.3. This test is very exiget! I most cases it will fail to solve your problem. For example you ca t apply it to the Riema series α. Theorem 4.4. [Raabe-Duhamel test] We suppose 0, a > 0.. If α R, the (i α > a coverges ; (ii α < a diverges. 2. Same coclusios if α R, a + a α + o (, a + a α ( l + o. l Proof. For. : if α > (resp. <, cosider α ], α[ (resp. ]α, [. To exploit the hypothesis, it s relevat to cosider the sequece b l( α a. Oe way to study such a sequece, cosiderig the l ad the ratio hypothesis, is to cosider the associated telescopig series u b + b : 5

6 ( u α l + + l a + ( a α l + ( + l α ( + o α ( O 2 α ( + o + O ( { α + o ( } 2 but (cf. defiitio of the Ladau otatios, o(/ 2 o(/ 2, (/o(/ o(/ 2 ad of course o(/ 2 O(/ 2, so (we also use the Mikowski iequality { α ( 2 ( } + o O 2. As we also have a, O(O(a O(a ad O(a / o(a (cf. / 0, we fially obtai ( u α α + o α α. Thus for (i, α α < 0 implies u, which meas b, which meas α a 0, which meas a o( α which gives the result (α >. For (ii, α α > 0 gives us α a +, so α O(a, which leads to the result (α <. For 2. : same proof, usig this time b l( l α (a. ( 2 (2! Exercise 4.5. Cosiderig a 2 2 (! 2, prove that the first Raabe test fails (α i the hyporhesis of., but ot the secod (α 0 i the hypothesis of Further results Theorem 5.. [Leibiz criterio] Suppose a ( b with (b a decreasig sequece which teds to zero. The. a coverges ; 2. if we ote S its sum, S 0 ; 3., r (a a + b +. Proof. (s 2 (a is decreasig, (s 2+ (a is icreasig ad s 2+ (a s 2 (a 0. Hece there exists S such that s 2+ (a S s 2 (a. As a cosequece of these iequalities, we have r (a S s (a s (a s + (a b +. For 2., just use S s 2 (a. Example 5.2. ( coverges. Lets s calculate its limit : we write ( k k with k k0 0 ( t k dt ( 0 ( t k dt k0 ( t α 0 + t dt 0 t dt ( t dt l 2 α + t 6

7 Fially ( l 2 Exercise 5.3. Prove that we ca apply the Leibiz criterio to r (a with a ( l. Defiitio 5.4. Let (a 0 ad (b 0 two sequeces. The Cauchy product of a ad b, oted ( a ( b, is the series c, with c a k b k. k0 Theorem 5.5. Suppose a ad b coverge ad ote A, B the sums of a, b. The c coverge ad its sum is AB. Proof. We write Let s prove α s (c hk i k a i b k i k0 i0 i a i i0 h0 b h i0 ki a i b k i a i s i (b i0 a i (B r i (b s (ab α }{{} i0 AB a i r i (b 0. For ɛ > 0, N sucht that N, r (b i0 0. We ote A the sum of a. The α N a i r i (b + Aɛ. }{{} N i0 But a 0, so N sucht that N, a ɛ. Hece. N + N, N N α (K + Aɛ N with K r i (b. i0 Propositio 5.6. [Abel s summatio by parts formula] Give to sequeces (a ad (b, we have the followig formulas p, q : q q (i a (b b (a a + b + a q+ b q a p+ b p (ii p+ q p+ p+ p+ q a b (a a + s (b + a q+ s q (b a p+ s p (b Proof. First, (ii is just (i applied to s (b istead of b. For (i : 7

8 q a (b b p+ q q a b a b p+ q p+ q p+ q p+ p+ q a b a + b a b p q p+ a + b a p+ b p + a q+ b q (a a + b + a q+ b q a p+ b p. Example 5.7. Let u cos(θ α. If α >, u O( α u coverges. If α 0, u 0 u diverges. If α ]0, ], we already kow that u diverges if θ 0 (mod 2π, so we may assume e iθ. I order to apply Abel s formula (ii, we ote a α ad b cos(θ ad we have (cf. s 0 (b cos 0 q u k0 q (a a + s (b +a q+s q (b. }{{} v But s (b ( ( e R e ikθ i(+θ R e iθ e iθ ( e i(+θ/2 2i si(( + θ/2 R e iθ/2 2i si(θ/2 si(( + θ/2 R(e iθ/2 si(θ/2 cos(θ/2 si(( + θ/2 si(θ/2 s (b K ad (with f α : x x α : si(θ/2 ( ( v K α ( + α K ( α + α, ( + α ( f α( α + α O ( Fially v O( (α+ ad v coverges (cf. α+ >. But a q+ 0, so the Abel s formula proves the covergece of u. 8

9 We fiish with the Fubii s theorem for double series : Theorem 5.8. Suppose (a m, C N N is such that for all m, a m, coverges to a limit oted σ m ad that σ m coverges to a limit oted Σ. The (i for all, m a m, coverges to a limit oted σ, (ii σ coverges, (iii m a m, m a,m (oted m, a m,. Proof. (i : For 0 N, we have for all M N M m0 a 0,m M m0 σ m Σ, so we have the result. (ii : For all N N N 0 σ N m0 0 a M N m, lim M m0 0 a m, N m0 0 a m, Σ, ad thus N 0 σ Σ ad N 0 a m, σ m, so M which is eough to coclude. (iii : First, both members of the equality exist : we ote S m a m, ad S m a m, so that S m σ m ad S σ imply the covergece of S m ad S. Let ɛ > 0. We have for all (M, N N N M m0 S m M m0 0 a m, M 0 m0 a m, N M 0 m0 a m, + M N+ m0 a m, where, because σ coverges, there exists N ɛ N such that for all N N ɛ N+ J m0 a m, N+ M m0 a m, N+ σ ɛ, ad where, because σ m coverges, there exists M ɛ N such that for all M M ɛ N 0 S N M 0 m0 a m, N 0 m M+ a m, N m M+ 0 a m, m M+ σ m ɛ. Hece for all N N ɛ ad M M ɛ we have which leads to the result. N 0 S M m0 S m 2ɛ. Remark 5.9. I fact (iii is a particular case of the double-limit theorem you ll see i ch2. The trick is to cosider E {x i } i N { } R with x i x ad to defie f m C E by f m (x i i 0 a m, for all i N { }. We have m, f m (x i xi x f m (x a m, ; ormal covergece : x E, f m (x σ m. Hece, settig g m 0 f m C E, lim xi x g(x i g(x, which exactly says that S coverges, ad that the limit is m S m. This theorem ca be very usefull for the theory of power series - see ch3. 9