Principles of Mathematical Analysis

Transcription

1 Ciro Uiversity Fculty of Scieces Deprtmet of Mthemtics Priciples of Mthemticl Alysis M 232 Mostf SABRI

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3 Cotets Locl Study of Fuctios. Remiders Tylor-Youg Formul Estimtio of the remider Exercises Numericl Sequeces 7. Limits of umericl sequeces Bsic properties of R Limits of rel sequeces Some stdrd limits Subsequeces d the Bolzo-Weierstrss Theorem Cuchy criterio Recursive sequeces Exercises Numericl Series Geerlities Series with oegtive terms Geerl series Summtio by pckets Rerrgemets Double series Cuchy product Exercises Sequeces d Series of Fuctios Poitwise covergece Uiform covergece Coservtio of properties by uiform covergece Uiform cotiuity Approximtio theorems Exercises Power Series Power series d rdius of covergece Properties of the sum fuctio Problems o the boudry Usul fuctios

4 iv Cotets 4.5 Alytic fuctios Power series d differetil equtios Exercises Improper Itegrls 6 5. Geerlities Bsic properties of improper itegrls Itegrls of oegtive fuctios Absolute covergece Cuchy criterio Exercises Bibliogrphy 69

5 Locl Study of Fuctios. Remiders Lemm.. Suppose tht f : ], b[ R ttis mximum or miimum t c ], b[. If f is differetible t c, the f (c) =. Proof. Suppose f hs mximum t c d let g(x) = f(x) f(c) x c. The if x > c, we hve g(x), hece f (c) = lim x c g(x), if x < c, we hve g(x), hece f (c) = lim x c g(x). Hece, f (c) =. If f hs miimum t c, the f hs mximum t c, hece f (c) = d thus f (c) =. Theorem.2 (Rolle s Theorem). Let f : [, b] R be cotiuous o [, b] d differetible o ], b[. If f() = f(b) =, the there exists c ], b[ such tht f (c) =. Proof. Sice f is cotiuous o [, b], f ttis its mximum t poit c [, b]. If f(c) =, the f(x) = for ll x (becuse f(x) f(c) ), hece f is ideticlly zero d the theorem is cler. If f(c) >, the c, b, moreover f hs mximum or miimum t c. Hece, f (c) = by the previous lemm. Theorem.3 (Me Vlue Theorem). Let f : [, b] R be cotiuous o [, b] d differetible o ], b[. The there exists c ], b[ such tht f(b) f() = f (c)(b ). Proof. Let g(x) = (b )f(x)+(f() f(b))x+f(b) bf(). The g (x) = (b )f (x)+ f() f(b). Moreover, g(b) = g() =, so by Rolle s Theorem, there exists c ], b[ such tht g (c) =, i.e. f(b) f() = f (c)(b ). Corollry.4. Let f : [, b] R be cotiuous o [, b] d differetible o ], b[. The (i) f is costt iff f =, (ii) f is icresig (resp. decresig) iff f (resp. f ), (iii) if f > (resp. f < ), the f is strictly icresig (resp. strictly decresig). Proof. If f = C d x ], b[, the f(x+h) f(x) h = C C h =, so f (x) =. Coversely, if f (x) = for ll x ], b[, let y >, the by Theorem.3, we my fid c ], y[ such tht f(y) f() = (y )f (c) =. Hece, f(y) = f() for ll y [, b] d f is costt. If f is icresig, the f(x+h) f(x) h, hece f (x). Coversely, if f, the give x < y b, we hve f(y) f(x) = (y x)f (c) (if f >, we lso hve f(y) f(x) > ). The cse of decresig fuctios is similr.

6 2 Locl Study of Fuctios.2 Tylor-Youg Formul Defiitio.5 (Ldu ottio). Let f, g be two fuctios defied er x R. Assume g(x) for x x. We sy tht. f = O(g) er x if there exists M > such tht f(x) M g(x) er x, 2. f = o(g) er x if lim x x f(x) g(x) =, 3. f g er x if lim x x f(x) g(x) =. Whe f = O(g), we sy tht f is domited by g, whe f = o(g), we sy tht f is egligible i frot of g, whe f g, we sy tht f is equivlet to g. f(x) f(x Recll tht f is differetible t x iff lim ) x x x x exists; i this cse we hve f(x) f(x by defiitio lim ) x x x x = f (x ). Hece, f is differetible t x iff f(x) = f(x ) + f (x )(x x ) + o(x x ) er x. ( ) This mes tht x f(x ) + f (x )(x x ) is the best lier pproximtio of f er x. The im of the Tylor-Youg formul is to give the best polyomil pproximtio of f er x. Lemm.6. Suppose g is differetible er d stisfies g (h) = o(h k ) er. The g(h) g() = o(h k+ ) er. Proof. Let ε >. Sice g (h)/h k s h, we my fid h > such tht g (h) ε h k for h ] h, h [ \ {}. Let h ] h, h [ \ {}. The by Theorem.3, we my fid x betwee d h such tht g(h) g() = g (x) h ε x k h ε h k+. Hece, g(h) g() h k+ s h. Theorem.7 (Tylor-Youg formul). Let f be defied er x d ssume f is differetible times t x. The er x we hve f(x) = f(x ) + f (x )(x x ) + f (x ) 2! (x x ) f () (x ) (x x ) + o ( (x x ) ).! Proof. We prove the result by iductio. Whe =, the theorem is true by ( ). Suppose the result is true for d pply it to f, which is differetible times t x. The tkig h = x x, we hve f (x + h) = j= (f ) (j) (x ) h j + o(h ) = j! j= Let g(h) = f(x + h) f(x ) k= f (k) (x ) k! h k. The g (h) = f (x + h) k= f (k) (x ) (k )! hk = f (x + h) j= Sice g() =, we hve g(h) = o(h ) by Lemm.6, s sserted. f (j+) (x ) h j + o(h ). j! f (j+) (x ) h j = o(h ). j!

7 .2. Tylor-Youg Formul 3 Corollry.8. The followig expsios re vlid er : e x = + x + x2 2! x! + o(x ) cos(x) = x2 2! + x4 x ( ) 4! (2)! + o(x2 ) si(x) = x x3 3! + x5 5! ( + x) = + x + x ( ) (2 + )! + o(x2+ ) ( ) x l( + x) = x x2 2 + x3 x ( )+ 3 + o(x ) ( ) ( + ) x + o(x )! Remrk.9. I Corollry.8, (i) we my replce o(x ), o(x 2 ) d o(x 2+ ) by O(x + ), O(x 2+2 ) d O(x 2+3 ), respectively. For exmple, e x = + x x! + x+ (+)! + o(x+ ). Sice x+ (+)! + o(x + ) = O(x + ), we hve the ssertio. (ii) We my remove o(x ), o(x 2 ) d o(x 2+ ) from the expsios, but replce = by. For exmple, l( + x) x d l( + x) x x2 2 er. Ideed, lim x l(+x) x = lim x x+o(x) x lim x ( + o(x) x 2 = lim x x x = lim x ( + o(x) x ) =, d lim x l(+x) x x2 2 ) =, sice o(x) s x (see Remrk 5 below). Remrk... If f = o(g), the f = O(g). 2. If f g, the f = O(g) d g = O(f). 3. If j k, the er, o(x j ) + o(x k ) = o(x k ). I prticulr, o(x) + o(x 2 ) = o(x). 4. x o(x k ) = o(x k+ ) d o(x k ) = x k o(). 5. If k, the lim x o(x k ) =. I prticulr, lim x o() =. 6. o(g) o(g) = o(g) d O(g) O(g) = O(g). 7. If c R, the c o(g) = o(g) d c O(g) = O(g) o(x2 ) x x2 2 Proof.. If f = o(g), the choosig ε =, we my fid δ > such tht f(x) g(x), i.e. f(x) g(x) if x x δ. 2. If f g, the choosig ε = 2, we my fid δ > such tht for x x δ, f(x) g(x) 2, i.e. 2 f(x) g(x) 3 2, so 2 f(x) g(x) 3 2 d thus f(x) 3 2 g(x) d g(x) 2 f(x). 3. If f/x j d g/x k, the (f + g)/x k = (f/x j )x j k + g/x k s x. 4. If f/x k, the xf/x k+. By iductio we get o(x k ) = x k o(). 5. If g = o(x k ), the lim x g(x) = lim x (g(x)/x k )x k =. 6. If f /g d f 2 /g, the (f f 2 )/g. If f C g d f 2 C 2 g er x, the f f 2 f + f 2 (C + C 2 ) g er x. 7. If f/g, the cf/g. If f C g, the cf C g.. O(g) d o(g) re sets of fuctios, so the equlity f = O(g) is ot very precise (it should be f O(g) d f o(g)). This is why we get some strge rules i Remrk.. Note lso tht the equlity sig is ot symmetric: = o(g), but o(g). Still, these ottios re very helpful i prctice. =

8 4 Locl Study of Fuctios cos(x) Exmple.. () lim x x si(x) = 2. Ideed, cos(x) = x2 /2 + o(x 2 ) d x si(x) = x 2 + o(x 2 ), hece cos(x) x si(x) = /2+o() +o() /2. (2) lim x e x 2 +x+ x 2 = 3 4. Ideed, + x = +x/2 x 2 /8+o(x 2 ), hece e x 2 + x+ = ( + x + x 2 /2 + o(x 2 )) (2 + x x 2 /4 + o(x 2 )) + = 3 4 x2 + o(x 2 ). (3) Expd f(x) = x2 si(x) +x to the order 5. Let g(x) = x 2 si(x) d h(x) = /( + x). The g(x) = x 3 x 5 /6 + o(x 5 ) d h(x) = x + x 2 x 3 + x 4 x 5 + o(x 5 ). Hece, usig remrks 3 d 4, f(x) = g(x)h(x) = (x 3 x 5 /6 + o(x 5 ))( x + x 2 x 3 + x 4 x 5 + o(x 5 )) = x 3 x 4 + 5x 5 /6 + o(x 5 )..3 Estimtio of the remider Theorem.2 (Tylor-Lgrge formul). Let f : [, b] R be cotiuous o [, b] d differetible ( + ) times o ], b[. The there exists c ], b[ such tht f(b) = f() + f ()(b ) + f () 2! Proof. Let g(x) = f(x) f(b) + i= (b x) i i! f (i) (x). The g (x) = f (x) i= Let λ = g() (+)! (b ) + (b x) i f (i) (x) + (i )! (b ) f () () (b ) + f (+) (c)! ( + )! (b )+. (b x) i f (i+) (x) = f (+) (b x) (x). i!! i= d h(x) = g(x) + λ (b x)+ (+)!. The h() = h(b) =, so by Rolle s Theorem, we my fid c ], b[ such tht = h (c) = (b c)! (f (+) (c) λ). Hece, λ = f (+) (c). Isertig this i the equtio h() =, we obti Tylor s formul. Theorem.3 (Tylor formul with itegrl remider). Let f : [, b] R be cotiuously differetible ( + ) times o ], b[. The f(b) = f() + (b )f () (b ) b f () () +! (b t) f (+) (t) dt.! Proof. We prove the result by iductio. For =, the theorem is true by the fudmetl theorem of itegrtio (f(b) = f() + b f (t) dt). Now itegrte by prts with u = f (+) (t) d v = (b t)+ (+)! to get b (b t)! f (+) (t) dt = (b )+ (+)! f (+) () + b (b t) + (+)! f (+2) (t) dt. Hece, if the result is true for, it is lso true for +..4 Exercises. Expd f(x) = to the order 2 er. +x+ x Expd f(x) = e x+ to the order 5 er. 3. Let g(x) = cos x o ] π 2, π 2 [. Expd g to the order 4 er. 4. Expd t x to the order 5 er.

9 .4. Exercises 5 5. Expd f(x) = x( + cos x) 2 t x to the order 3 er. Deduce fuctio which is equivlet to f. 6. Clculte the followig limits. Here, b >. x( + cos x) 2 t x lim x 2x si x t x x b x lim x x ( lim x x ) l( + x) ( x + b x ) /x lim x 2 7. Give exmple of fuctios f, g, h such tht f g er, but the reltio h+f h+g is ot true. Show tht if h/g hs limit, the h + f h + g.

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11 Chpter Numericl Sequeces. Limits of umericl sequeces Defiitio.. A umericl sequece is mp u : N K, where K = R or C. We deote u := u(). The mp itself is deoted by (u ) N. If K = R, (u ) N is rel sequece. If K = C, (u ) N is complex sequece. More geerlly, sequece my strt from idex k N. I this cse, we deote it by (u ) k. For simplicity, we shll ofte deote sequeces by (u ). Defiitio.2. We sy tht sequece (u ) is bouded if M > : u M N. Defiitio.3. We sy tht l C is limit for umericl sequece (u ) if ε > N : = u l ε. If (u ) hs limit, we sy tht (u ) is coverget. Otherwise, we sy tht (u ) is diverget. Lemm.4 (Uiqueess of the limit). Ay sequece hs t most oe limit. Proof. Suppose o the cotrry tht umericl sequece (u ) hs two limits l l d choose ε := l l 3 >. The there exists idex such tht u l ε, d idex 2 such tht u l ε 2. Let 3 = mx(, 2 ). The l l = (l u 3 ) + (u 3 l ) l u 3 + u 3 l 2ε = 2 3 l l. This cotrdictio shows tht l = l. The uique limit of sequece, whe it exists, is deoted by lim u, or more simply by lim u. If (u ) hs limit l, we sy it coverges to l d deote u l. Remrk.5. Give two sequeces u d v, if : u = v, d if u hs limit l C, the v coverges to the sme l. I other words, if sequece coverges, d if we chge fiite umber of its terms, the this does ot chge the limit. Lemm.6. Ay coverget sequece is bouded. Proof. Suppose (u ) coverges to l C. There there exists N such tht u l. Hece, u u l + l + l. Tke M := mx{ u, u,..., u, + l }. The u M for ll, hece (u ) is bouded.

12 8 Chpter. Numericl Sequeces Lemm.7. If lim u = l C d lim v = l C, the for y c C, (i) (iii) lim cu = cl, (ii) lim (u + v ) = l + l, + + lim u v = ll. + Proof. (i) Give ε >, there exists such tht = cu cl = c u l c ε ε, where ε = ε/ c if c d ε = if c =. (ii) Give ε >, we hve (u + v ) (l + l ) = (u l) + (v l ) u l + v l. We kow there exist, 2 N such tht u l ε 2 d v l ε 2 2. Tke 3 = mx(, 2 ). The (u + v ) (l + l ) ε 3. (iii) Give ε >, we hve u v ll = u v + u l u l ll = u (v l ) + (u l)l u v l + u l l. Sice (u ) coverges, it is bouded, so there exists M > such tht u M for ll. O the other hd, we my fid, 2 N such tht v l ε 2M for d u l ε for 2, where ε = ε 2 l if l d ε = if l =. Tke 3 = mx(, 2 ). The u v ll ε 3. Lemm.8. Suppose u l. The there exists k such tht u for ll k. Moreover, the sequece (/u ) k coverges to /l. Proof. Give ε >, we hve u l = u l lu. We strt by givig lower boud o the deomitor. We hve u l u l. Choosig ε = l 2, we kow tht u l ε strtig from idex k. Hece, u l ε = ε d we get = u l u l lu l ε u l for ll k. But we my fid k 2 such tht u l ε for ll k 2, where ε = l ε ε. Hece, u l ε for ll mx(, 2 ). Lemm.9. If (u ) is bouded d (v ) coverges to, the (u v ) coverges to. Proof. Let ε >. We my fid M > such tht u M for ll. O the other hd, choosig ε = ε M, we my fid such tht v = v ε for ll. Hece, u v Mε = ε for ll. Defiitio.. Let z K, where K = R or C, let B r (z ) := {z K : z z < r} d let B r (z ) = {z K : z z r}. If K = R, the B r (z ) = ]z r, z + r[ d B r (z ) = [z r, z + r]. If K = C, the B r (z ) is ope disk of rdius r >, cetered t z, while B r (z ) is closed disk. Defiitio.. Let f : B r (z ) C d p B r (z ). We sy tht lim z p f(z) = l if ε > δ > : z B r (z ), z p δ = f(z) l ε.

13 .2. Bsic properties of R 9 Lemm.2. Let f : B r (z ) C d p B r (z ). The lim f(z) = l ( {u } B r (z ) : u p = f(u ) l ). z p Proof. Let ε > d suppose lim z p f(z) = l. The we my fid δ > such tht f(z) l ε for y z B r (z ) stisfyig z p δ. But if u p, the such tht u p δ for ll. Hece, f(u ) l ε for ll d thus f(u ) l. Coversely, suppose tht lim z p f(z) l. The ε > δ > : z B r (z ), z p δ but f(z) l > ε. Hece, for δ = 2 we my fid u B r (z ) such tht u p 2 but f(u ) l > ε. Hece, u p but f(u ) l. The ssertio follows by cotrpositio. Exmple.3. lim x si( x ) does ot exist. Ideed, if x 2 = (2+)π, the x but si( x ) = si( (2+)π 2 ) = ( ) hs o limit. Hece, si( x ) hs o limit by Lemm.2. Defiitio.4. We sy tht f : B r (z ) C is cotiuous t p B r (z ) if ε > δ > : z B r (z ), z p < δ = f(z) f(p) ε. Lemm.5. Let f : B r (z ) C d p B r (z ). {u } B r (z ), if u p, the f(u ) f(p). The f is cotiuous t p iff Proof. Comprig Defiitio. d Defiitio.4, we see tht f is cotiuous t p iff lim z p f(z) = f(p). Hece, the ssertio follows from Lemm.2. Exmple.6. If u l, the u l, sice x x is cotiuous fuctio..2 Bsic properties of R Defiitio.7. We sy tht r R is upper boud for set A R if r A. We sy tht r R is lest upper boud for A R if r is upper boud for A, y s < r is ot upper boud for A. The lest upper boud is lso clled the supremum of A d is deoted by r = sup(a). Theorem.8. R hs the lest-upper-boud property. Tht is, if oempty A R hs upper boud, the it hs lest upper boud. Proof. Admitted. Theorem.9. Give x > d y R, there exists N such tht x y. We refer to this s the rchimede property of the rels. Proof. Suppose o the cotrry tht x < y for ll N. The A := {x : N} is oempty set of rel umbers which hs upper boud y. Hece, it hs lest upper boud r = sup(a). Sice r x < r, r x is ot upper boud for A, so there exists m N such tht r x < mx. Hece, r < (m + )x, which is impossible, sice r is upper boud for A. This cotrdictio proves the theorem.

14 Chpter. Numericl Sequeces Corollry.2. Give x > d y R, there exists N such tht x y. Proof. Let x = + h, h >. The by the biomil theorem, ( + h) + h. By the rchimede property, we my fid such tht h y. Hece, x y. Lemm.2. Give x R d >, there exists uique Z such tht x < ( + ). Proof. Uiqueess : If d stisfy the ssertio, the x < + = < = } x < + = < = = =. Existece : By the rchimede property, we my fid, 2 N such tht x d 2 x. Thus, A := {k Z : k x} is oempty subset of Z ( 2 A) which hs upper boud. Hece, A hs gretest elemet A. Sice + / A, we get x < ( + ). The cse = i the previous lemm is of specil iterest : Defiitio.22. If x R, the uique Z stisfyig x < + is clled the iteger prt of x, d deoted by [x]. The rel x [x] is clled the frctiol prt of x, d is deoted by {x}..3 Limits of rel sequeces We re ow iterested i certi otios tht use the turl order of the rels. Let us first ote the followig result. Lemm.23. (i) If rel sequece (u ) hs limit l, the l R. (ii) A complex sequece (u ) coverges iff the sequeces (Re u ) d (Im u ) coverge. Proof. (i) Put l = + ib, where, b R. The for y ε > we hve u l ε strtig from idex. Thus, b (u ) 2 + b 2 = u l ε for y ε >, which mes tht b =. (ii) If (Re u ) d (Im u ) coverge, the (u ) = (Re u +Im u ) coverge by Lemm.7. Coversely, suppose u l C d put l = + ib,, b R. The Re u (Re u ) 2 + (Im u b) 2 = u l, so the covergece of (u ) to l implies the covergece of (Re u ) to. Similrly, the sequece (Im u ) coverges to b. Lemm.24 (Pssig to the limit). Let u d v be two rel coverget sequeces. If there exists k N such tht u v for ll k, the lim u lim v. I prticulr, tkig the costt sequece v, if u for ll, the lim u.. This follows from the well-orderig priciple, which sys tht y oempty subset of N hs lest elemet.

15 .3. Limits of rel sequeces Proof. Suppose u l d v l 2. By hypothesis, we hve u v = u v for ll lrge. Hece, l l 2 = lim(u v ) = lim u v = l l 2. Lemm.25. Let (u ) be complex sequece d l C. If there exists rel sequece (v ) covergig to such tht u l v for ll, the u l. Proof. Let ε >. Sice v, there exists such tht v ε for ll, d thus u l v ε for ll. Exmple.26. Let f be bouded fuctio defied o X R. If M = sup X f, the we my fid sequece (x ) of elemets of X such tht f(x ) M. Ideed, M is ot upper boud, so there exists x X such tht M < f(x ). But M is upper boud, so f(x ) M. Hece, < f(x ) M d thus f(x ) M. Hece, f(x ) M by Lemm.25. Corollry.27 (Sdwich Theorem). Let (u ), (v ) d (w ) be rel sequeces. If there exists k such tht u v w for ll k, d if lim u = lim w = l R, the (v ) coverges, d lim v = l. Proof. We hve hypothesis v u w u for y k. Sice lim(w u ) =, the sequece (v w ) teds to by Lemm.25. Sice v = u + (v u ), we hve lim v = l + = l. Defiitio.28. We sy tht rel sequece (u ) teds to + if M R N : = u M, teds to if ( u ) teds to +, i.e. if M R N : = u M. Note tht if u d v re rel sequeces, d if there exists k N such tht u v for ll k, the ( u + = v + ), d ( v = u ). This is cler from the defiitios, d my be see s ppedix to the sdwich theorem. Lemm.29. Let (u ) be rel sequece. The u + iff k N : u > k, d the sequece ( u ) k teds to. Proof. Suppose u + d let ε >. The we my fid k such tht u /ε for ll k. Thus, for k, we hve u > d < u ε. Thus, u. Coversely, let M >. If u, the we my fid k such tht < u M, sice u >. Hece, u M d u +. Defiitio.3. We sy tht rel sequece (u ) is bouded from bove if there exists M R such tht u M for ll, bouded from below if there exists m R such tht u m for ll, icresig if u + u for ll, decresig if u + u for ll, mootoe if it is either icresig or decresig. We sy it is strictly icresig, strictly decresig or strictly mootoe if the correspodig iequlity is strict.

16 2 Chpter. Numericl Sequeces Theorem.3. Let u be icresig rel sequece.. If it is bouded from bove, the it coverges to l = sup{u : N}. 2. If it is ot bouded from bove, the it teds to +. If v is decresig d bouded from below, the it coverges to l = if{v : N}. Proof.. If u is bouded from bove, the A := {u : N} hs lest upper boud l by Theorem.8. Let ε >. The l ε is ot upper boud, so N such tht l ε < u. Sice u is icresig d bouded from bove by l, we get hece u l. = l ε < u u l, 2. If u is ot bouded from bove, the give M >, we my fid such tht u > M. Sice u is icresig, we thus hve u u > M for ll. Thus, u +. Filly, if u = v, the u is icresig d bouded from bove, hece coverges to sup u = sup( v) = if v, so tht v = u coverges to sup u = if v. Defiitio.32. We sy tht two rel sequeces u d v re djcet if N, u v, u is icresig d v is decresig, (v u ) teds to. Lemm.33. If u d v re two djcet sequeces, the they coverge to commo limit l stisfyig u l v for ll. Proof. We hve u v v for ll, so u is icresig d bouded from bove. Hece, u coverges d u lim u by Theorem.3. Similrly, v is decresig d bouded from below, hece coverges d we hve v lim v. Filly, v = u+(v u) d lim(v u) =, so lim u = lim v. Defiitio.34. We sy tht (I ) is sequece of ested itervls if ech I is itervl of R d I + I. Corollry.35 (The Nested Itervl Theorem). If ([, b ]) is sequece of ested closed itervls whose legth teds to, the the set N [, b ] cosists of sigle poit. Proof. By hypothesis, the sequeces ( ) d (b ) re djcet, so they possess commo limit l stisfyig l b for ll. The poit l thus belogs to ech [, b ], hece the itersectio of these itervls cotis l d is oempty. O the other hd, if x N [, b ], the x b for ll, so pssig to the limit we get l = lim x lim b = l. Hece, x = l d N [, b ] = l..4 Some stdrd limits Defiitio.36. Let (u ) d (v ) be two sequeces of umbers. We sy tht. u = O(v ) if there exists N d C > such tht u C v for ll. 2. u = o(v ) if there exists sequece w such tht u = w v. 3. u v if there exists sequece w such tht u = w v. Note tht if v is ever, the u = o(v ) iff u v d u v iff u v.

17 .4. Some stdrd limits 3 Remrk.37. If u = o(v ), the u = O(v ). v = O(u ). This is proved s i Chpter. If u v, the u = O(v ) d Theorem.38. () If α >, the lim α =. (b) If C d <, the lim =. (c) If p >, the lim p =. (d) lim =. (e) If > d α R, the lim α =. (f) If >, the lim = +. (g) For y α, β >, we hve lim (l ) α β =. Proof. () Let u =. Give ε >, we use the rchimede property to fid α N such tht ε /α. The u α ε for y. (b) If =, this is cler, otherwise >, so give ε > we my use Corollry.2 to fid such tht ( ) ε. Hece, ε for y. (c) If p =, this is cler. If p >, put x = p. The x > d by the biomil theorem, + x ( + x ) = p. Thus, < x p d x. If < p <, we obti the ssertio by cosiderig q = p. (d) Put x =. The x d = (+x ) ( ) 2 x 2. Hece, x for 2. Hece, x. 2 (e) Put = + h, h >. Let k be iteger such tht k > α d k >. The for > 2k, we hve = > 2 + k, so by the biomil formul, = ( + h) > ( ) h k = k ( ) ( k + ) h k > k h k k! 2 k k!. Thus, < α < 2k k! h k α k for y > 2k. Sice α k <, we hve α k. (f) This follows from (e) by tkig α = d pplyig Lemm.29. (g) We first show tht (l ) α C β/2 for some C >. Give γ >, let f(x) = γ x γ l x. The f (x) = x γ x = xγ x for y x. Furthermore, f() = γ >. Hece, f(x) > for y x. Tkig γ = β 2α, we thus get the required iequlity for y N. The vlue of the limit ow follows from (). Lemm.39 (Deciml pproximtio of rel umber). For y x R, the sequece u = [ x] coverges to x. I prticulr, y rel umber is the limit of sequece of rtiols. We sy tht Q is dese i R. Proof. By defiitio, [ x] x < [ x] +, hece u x < u +. Thus, x u < d thus u x. Exmple.4. If x = 2, the u =, u =.4, u 2 =.4 d u 3 =.44.

18 4 Chpter. Numericl Sequeces.5 Subsequeces d the Bolzo-Weierstrss Theorem Defiitio.4. A sequece (v ) N is sid to be subsequece of (u ) N if there exists strictly icresig mp ϕ : N N such tht v = u ϕ() for ll N. Exmple.42. (u + ) N, (u 2 ) N d (u 2+ ) N re subsequeces of (u ) N. Note tht if ϕ : N N is strictly icresig, the we see by iductio tht ϕ() N. Lemm.43. If u l, the y subsequece (v ) of (u ) lso coverges to l. Proof. Give ε >, we my fid such tht u l ε for y. But v = u ϕ() for some strictly icresig ϕ : N N. Hece, ϕ() d = ϕ() = u ϕ() l ε. Exmple.44.. The sequece ( ) is diverget, sice the subsequece (u 2 ) teds to, while the subsequece (u 2+ ) teds to. 2. The sequece u = cos(π/4) is diverget, sice the subsequece u 4 = ( ) diverges. Lemm.45. If the subsequeces (u 2 ) d (u 2+ ) of u coverge to commo limit l, the u lso coverges to l. Proof. Give ε >, we my fid, 2 N such tht = u 2 l ε d 2 = u 2+ l ε. Tke = mx(2, ). The for, we hve u l ε, hece u l. Lemm.46 (cf. [2]). Ay rel sequece hs mootoe subsequece. Proof. Let (u ) be rel sequece. Cll u domit if u > u for ll >. If (u ) hs ifiite umber of domit terms with idices < <..., the the subsequece (u j ) j N is decresig, sice u k > u k+. So the result is true i this cse. If (u ) hs fiite umber of domit terms, let u N be the lst oe d put m = N +. The u m is ot domit, so there exists m 2 > m such tht u m u m2. Similrly, u m2 is ot domit sice m 2 > N, so there exists m 3 > m 2 such tht u m2 u m3. By iductio, there exists subsequece u m u m2 u m3.... This completes the proof. Corollry.47 (Bolzo-Weierstrss Theorem). Ay bouded sequece hs coverget subsequece. Proof.. Let u be rel bouded sequece. By Lemm.46, it hs mootoe subsequece v, which is of course bouded too. Hece, v coverges by Theorem Let u be complex bouded sequece d put u = x + iy, with x, y R. Sice x x 2 + y 2 = u, the sequece (x ) is rel d bouded, so it hs coverget subsequece (x ϕ() ) with limit. Similrly, y ϕ() u ϕ(), so (y ϕ() ) hs coverget subsequece (y ϕ(ψ()) ) with limit b. Filly, (x ϕ(ψ()) ) is subsequece of (x ϕ() ), so x ϕ(ψ()) by Lemm.43. Thus, u ϕ(ψ()) = x ϕ(ψ()) + iy ϕ(ψ()) + ib.

19 .6. Cuchy criterio 5.6 Cuchy criterio Defiitio.48. We sy tht (u ) is Cuchy sequece if ε > N : p, q = u p u q ε. Remrk.49. Equivletly, (u ) is Cuchy sequece if ε > N : r, s = u r u r+s ε. Ideed, to go from the first defiitio to the secod, tke p = r d q = r + s. For the coverse, tke r = mi(p, q) d s = mx(p, q) mi(p, q). Theorem.5. A sequece coverges iff it is Cuchy sequece. Proof. Suppose u l d let ε >. Choose such tht u l ε/2 for ll. The for p, q we hve u p u q = (u p l) (u q l) u p l + u q l ε. Coversely, suppose (u ) is Cuchy. We first show tht (u ) is bouded. Tkig ε =, we my fid such tht u p u q for ll p, q. I prticulr, u + u for y. Put M = mx{ u,..., u, + u }. The u M for ll, hece (u ) is bouded. It follows from the Bolzo-Weierstrss theorem tht (u ) hs coverget subsequece (u ϕ() ) with limit l. So such tht u ϕ() l ε/2 for ll. But (u ) is Cuchy, so 2 such tht u p u q ε/2 for ll p, q 2. Sice ϕ(), we thus hve i prticulr u u ϕ() ε/2 for ll 2. Hece, if 3 = mx(, 2 ), we hve hece (u ) coverges to l. 3 = u l u u ϕ() + u ϕ() l ε,.7 Recursive sequeces I the followig, u : N K, where K = R or C..7. Arithmetico-geometric sequeces We sy tht (u ) is rithmetico-geometric sequece if there exist, b K such tht the followig recurrece reltio is stisfied : N : u + = u + b. If =, we hve rithmetic sequece, hece Suppose, we show tht N : u = u + b. N : u = (u r) + r, where r = b. Ideed, put v = u r. The v + = u +b r = v +r+b r = v. Hece, (v ) is geometric sequece with commo rtio. Hece, u = v + r = v + r = (u r) + r.

20 6 Chpter. Numericl Sequeces If u = r, the sequece is thus costt d coverges to r. If <, we lso hve u r. If u r d >, the u u r r +, so (u ) is diverget (if (u ) ws coverget, ( u ) would lso be coverget). We will see i Exercise 7 tht if =, the ( ) coverges iff =. Coclusio : if <, the u r. If, the (u ) coverges iff (u = r) or ( = d b = ), i which cse (u ) is costt..7.2 Lier recursive sequeces of order 2 We sy tht (u ) is lier recursive sequece of order 2 if N : u +2 = u + + bu ( ) for some, b K. Lemm.5. Let P (X) = X 2 X b. () If P hs two distict roots r d r 2, the the solutios of ( ) re the sequeces (αr + βr 2 ) N, with α, β K. (b) If P hs double root r, the the solutios of ( ) re the sequeces ((α + β)r ) N, with α, β K. (c) If K = R d P hs o rel roots, there exist ρ > d θ / {kπ : k Z} such tht the solutios of ( ) tke the form (ρ (λ cos(θ) + µ si(θ))) N, with λ, µ R. Proof. Let E = {(u ) N : (u ) N stisfies ( )}. Usig elemetry lier lgebr, we see tht E is vector spce of dimesio 2 (the mp f : E K 2 give by f : (u ) N (u, u ) is isomorphism). Hece, there re exctly two lierly idepedet sequeces which stisfy ( ). I cse (), the sequeces (r ) d (r 2 ) both stisfy ( ), d they re ot proportiol. Hece, they form bsis for E, d this proves (). I cse (b), the sequece v = r stisfies ( ). Let us show tht w = r lso stisfies ( ). Sice P hs double root r, the discrimit = 2 + 4b is zero d we hve r = 2. Thus, w +2 = ( + 2)r r 2 = ( + 2)r (r + b) = ( + )r + + b(r ) + r + + 2br = w + + bw + ( b)r = w + + bw becuse r = 2 d =. Hece, (v ) d (w ) form bsis for E, d this proves (b). Filly, i cse (c), P hs roots i C of the form r ± = ρe ±iθ, with ρ > d θ / {kπ : k Z}. Sice the sequeces (r +) d (r ) both stisfy ( ), their lier combitio (ρ cos(θ)) d (ρ si(θ)) lso stisfy ( ). Moreover, they re rel d form bsis for E. This yields (c)..7.3 Geerl recursive sequeces of order We coclude this chpter with study of rel sequeces (u ) N stisfyig recurrece reltio of the form N : u + = f(u ), where f : I R is fuctio defied o itervl I R, with u I. We lso suppose tht I is stble by f, i.e. f(i) I, to esure tht the sequece is well defied (ideed, if sy u = f(u ) / I, the u 2 = f(u ) is udefied).

21 .7. Recursive sequeces 7 Defiitio.52. We sy tht itervl I R is closed if it hs the form I = [, b], I = [, + [, I = ], b] or I = R. Lemm.53. If I is closed, u I d u l R, the l I. Proof. This is trivil if I = R. If I = [, + [, the u, hece lim u by Lemm.24 d thus l I. The other cses re proved similrly. Lemm.54. Let f : I I be cotiuous fuctio o closed itervl I. sequece u + = f(u ), u I hs limit l, the l I d l = f(l). If the Thus, the limit of such sequece must be fixed poit of f. Proof. Sice u I d I is closed, the l I by Lemm.53. O the other hd, u l implies f(u ) f(l) by Lemm.5, hece u + f(l). But (u + ) N is subsequece of (u ) N, so u + l. The uiqueess of the limit ow shows tht l = f(l). Exmple.55. If u + = u 2 + 2, the u diverges, becuse the equtio x = f(x) hs o solutio. Ideed, x R, x < x Tkig x = u, we see tht (u ) is strictly icresig. Sice it diverges, it teds to +. Lemm.56. If there exists l I d k [, [ such tht the mp f : I I stisfies x I : f(x) l k x l, the give u I, the sequece u + = f(u ) coverges to l. Proof. We see by iductio tht u l k u l, so u l, sice k <. Exmple.57. Let u. We defie the sequece u + = 5u+3 u +5, whose terms re oegtive becuse the itervl R + is stble by f : x 5x+3 x+5. The equtio x = 5x+3 x+5 hs two roots ± 3. Sice ll the u re oegtive, the limit of u must be oegtive. Hece, if u coverges, it must ted to 3 by Lemm.54. Give x, the iequlity f(x) 3 = 5x + 3 x x + 5 proves tht u 3 by Lemm.56. = (x 3)(5 3) x + 5 ( ) x 3 Defiitio.58. A mp f : I R is cotrctio if it is k-lipschitz cotiuous o I, with k [, [. Tht is, if k [, [ such tht x, y I : f(x) f(y) k x y. Corollry.59. If l I is fixed poit of f d if f : I I is cotrctio, the the sequece coverges to l. Remrk.6. If f is differetible o I, d if f (x) k o I, where k [, [, the f is cotrctio o I by the me vlue theorem.

22 8 Chpter. Numericl Sequeces Exmple.6. Let u = d u + = 4 si u + 2. Sice the itervl [, ] is stble by f(x) = 4 si x + 2, this defies sequece of elemets i [, ]. The fuctio g(x) = x f(x) stisfies g() < d g() >. Sice it is cotiuous d strictly icresig, there must exist uique poit l such tht g(l) =, i.e. l = f(l). The fuctio f is 4 -Lipschitz cotiuous, sice it is differetible d f (x) 4. Hece, the sequece coverges to l. We c ctully estimte the speed of covergece : ( ) u ( ) u l l. 4 4 Sometimes the precedig criteri re ot sufficiet to coclude. We c the try to study the mootoicity of the sequece. For istce, if we show tht (u ) is icresig d bouded from bove, the we c coclude tht it coverges (Theorem.3). We hve the followig criteri. Lemm.62. If f(x) x x I, the the sequece u + = f(u ) is icresig. If f(x) x x I, the the sequece is decresig. Proof. We hve u + u = f(u ) u. Lemm.63. If f is icresig, the the sequece u + = f(u ) is mootoe. Proof. If u u, the u = f(u ) f(u ) = u 2 d by iductio, u u +. Similrly, if u u, the (u ) is decresig. Remrk.64. If f is icresig, we c use the fixed poits of f to boud the sequece. For istce, if f() = d u, the u + = f(u ) f() =. Exmple.65. Let u d u + = u 2 u +. Sice f(x) = x 2 x+ = (x 2 ) , we see tht R + is stble by f. The equtio f(x) = x hs uique solutio x =. The estimte f(x) = x x gives o iformtio. So we will rgue by mootoicity. We hve x R +, f(x) x, so the sequece (u ) is icresig. Hece, if u >, the sequece cot coverge to, the oly fixed poit of f, so it diverges to +. If u, the sequece is icresig d bouded bove by (use Remrk.64). Hece, the sequece coverges, d its limit must be. We coclude this chpter with importt theorem. I cotrst to Corollry.59, this theorem proves the existece of fixed poit uder certi ssumptios. Theorem.66 (The Fixed Poit Theorem 2, cf. [3]). If f : I I is cotrctio o closed itervl I, the f hs uique fixed poit l I. Moreover, for y c I, the sequece u = c, u + = f(u ) coverges to l. The speed of covergece my be estimted by u l k k u u, where k [, [ is the Lipschitz costt of f. 2. of Bch or Picrd.

23 .8. Exercises 9 Proof. For y x, y I we hve d thus x y x f(x) + f(x) f(y) + f(y) y x f(x) + k x y + f(y) y, x y f(x) x + f(y) y k. ( ) This lredy shows the uiqueess of the evetul fixed poit : if x d y re two fixed poits, the x y =. To prove its existece, let f = f f be the mp f composed times with itself, with the covetio f (x) = x. Let c I. We my write the sequece u = c, u + = f(u ) i the form (f (c)) N. Let us show tht this is Cuchy sequece. It will the follow tht it coverges to fixed poit l I by Lemm.54, which is the required result. We first ote by iductio tht f (x) f (y) k x y for y x, y I. Tkig x = f (c) d y = f m (c) i ( ), we thus get f (c) f m (c) f(f (c)) f (c) + f(f m (c)) f m (c) k = f (f(c)) f (c) + f m (f(c)) f m (c) k k f(c) c + k m f(c) c k = k + k m k u u. Sice k [, [, this teds to s, m +, hece (f (c)) N is Cuchy sequece. Keepig fixed d lettig m +, we obti the estimte o the speed of covergece..8 Exercises. Clculte the limits of the followig sequeces, if they exist : 3 ( 2) 3 + ( 2) ( + ) ( si )! ( ) / e si + ( ) ( 2 + ( ) )! si(). 2. Let u = (u ) N, v = (v ) N be two rel sequeces. Are the followig ssertios correct? Justify your swer. () If u is icresig d bouded from bove, the it coverges. (b) If u is bouded from bove d coverget, the it is icresig. (c) If u coverges to l R, the it is bouded. (d) If u is decresig d oegtive, the it coverges. (e) If u is decresig d strictly positive, the it coverges to strictly positive limit.

24 2 Chpter. Numericl Sequeces (f) If u diverges to d v diverges to +, the u + v coverges to. (g) If u teds to d the u re ozero, the /u teds to + or. (h) If u d v diverge to, the u + v diverges to. (i) If u coverges d v diverges to +, the uv diverges. (j) If u teds to, the uv teds to. (k) If u is icresig d v coverges to, the u v is icresig. (l) If u teds to + d v is bouded from below, the u + v teds to +. (m) If u coverges d v does ot coverge, the uv does ot coverge. () If u teds to + d v is bouded from below beyod certi idex by strictly positive rel umber, the uv teds to Let (u ), (v ) be two rel sequeces defied by u := +! + 2! ! d v := u +!. Show tht (u ) d (v ) re djcet. Deduce tht they coverge to commo limit. We will dmit tht this limit is the rel umber e. We would ow like to prove tht e / Q. For this, show tht we c write u =! with N d deduce the iequlity < e! < + for ll. Suppose tht e = p q Q d deduce cotrdictio by choosig = q i the previous iequlity. 4. Let u = (u ) N, v = (v ) N be two rel sequeces which do ot vish beyod certi idex. Are the followig sserios correct? Justify your swer. () If u v, the for ll k Z, we hve u k v k ; (b) If u v, the e u e v ; (c) If u v, the l u l v ; (d) If u = + o(), the e u e ; (e) If u v, the u v ; (f) If u v, the u v ; (g) If u = o(v ), the u = o(v 2 ). 5. Let d b be strictly positive, b, d let (u ) N d (v ) N be two sequeces defied by () Show tht for ll N, u v. u = d v = b u + = 2u v d v + = u + v. u + v 2 (2) Show tht (u ) d (v ) coverge d tht their limits re equl. (3) Usig the product u v, determie the vlue of this limit. (4) Applictio: give rtiol pproximtios to 2 d 3.

25 .8. Exercises 2 6. (Cesàro mes). Let (u ) be umericl sequece d (v ) the sequece of Cesàro mes defied by v := u k. () Show tht if u, the v. (2) Deduce tht if u l C the v l. k= (3) Does the covergece of v imply the covergece of u? (4) Applictio: Let (x ) be sequece with strictly positive terms. Suppose the sequece y = x + x coverges to l >. Show tht the sequece (z ) defied by z = x coverges to l. (5) Applictio: Determie the limit of the sequece ( 2) /. (6) Give exmple of sequece (x ) such tht x l but the limit of x + x does ot exist. 7. Let z be complex umber with z = such tht the sequece (z ) coverge. Determie the limit of this sequece. Hit. You my use Exercise Let (u ) be rel sequece such tht, for ll p N, Does the sequece (u ) coverge? lim (u +p u ) =. 9. Study the covergece of the followig sequeces, d give their limits if possible. u = d u + = 2u u d u + = u + 2 u + u = d u + = 2 u u = d u + = 3u 2 2u u, u d u +2 = 2u + u u =, u = d u +2 = 4u + 4u + 2 u = 3 d u + = u 3 exp(u ). Let (u ) be rel sequece d put S = u k d T = k= () Show tht if S coverges, the u. (2) Show tht if T coverges, the u teds to i the sese of Cesàro, i.e. k= u k. Hit. Express u k s fuctio of T j the use Exercise 6.. Let N d let f : [, ] R be the fuctio defied by f (x) := x + x 2 + x. () Fix N. Show tht f (x) is cotiuous, strictly icresig fuctio. (2) Deduce tht f hs exctly oe zero. We deote x [, ] the zero of f. (3) Fix x [, ]. Show tht the sequece (f (x)) is decresig. Deduce tht the sequece (x ) is icresig. (4) Show tht (x ) coverges. We deote its limit by l. k= u k k.

26 22 Chpter. Numericl Sequeces ( ) (5) Show tht for ll N, f 3 4 >. Deduce tht x < 3 4 d tht x s. (6) Coclude tht l 2 + l =. Wht is the vlue of l? 2. (Fekete s Lemm). Let (u ) be rel sequece stisfyig u +m u + u m for y, m N. Show tht ( u ) u teds to l = if R { }. 3. Let (u ) be oegtive, decresig sequece tht teds to. We put ( ) v := u k u. k= () Let p N. Show tht for ll p, we hve v p k= (u k u ). (b) Deduce tht if (v ) is bouded sequece, the the sequece S m = m k= u k coverges. (c) Study the coverse. 4. Let (u ) be rel sequece. We sy tht l R is limit poit of the sequece if there exists subsequece of (u ) which coverges to l. ) Wht re the limit poits of the sequece u = ( )? the sequece u = si( π 3 )? b) Show tht if sequece coverges, the it hs uique limit poit. Is the coverse true? c) Show tht every bouded rel sequece which hs uique limit poit is coverget. d) Applictio: Let (u ) be bouded rel sequece such tht u + u 2 2 coverges. Show tht (u ) coverges. e) Applictio: Let ( ) d (b ) be two rel sequeces such tht + b d e + e b 2. Show tht the two sequeces coverge.

27 Chpter 2 Numericl Series 2. Geerlities Defiitio 2.. Give sequece (u ), we cll = u = u + u +... series with geerl term u. We cll S k := k = u the prtil sums of the series. We sy tht the series coverges if the sequece S k coverges. I this cse, S = lim k S k = = u is clled the sum of the series. R = S S = k=+ u k is clled the remider of the series. Exmple 2.2. The geometric series λ coverges iff λ <. Ideed, if λ =, the series diverges. If λ, we hve S k = + λ λ k d λs k = λ λ k+. Hece, S k λs k = λ k+, i.e. S k = λk+ λ. We see tht (S k) coverges iff λ <. I this cse, = λ = lim k S k = λ. Remrks 2.3. If u d v coverge, the λ u + µ v := (λu + µv ) coverges. The coverse is clerly ot true. The covergece of series u is uchged if we modify fiite umber of u. Lemm 2.4. If u coverges, the u. Proof. If, we hve u = S S. Hece, S S = u S S =. The coverse is ot true. For exmple, u = l( + x) x with x =, we hve. However, usig the idetity which teds to s. S = k ( k + ) l = l( + ), k k= k= Lemm 2.5. If u coverges, the R. Proof. Assume S S. The R = S S S S =. Exmple 2.6. (+) coverges. Ideed, (+) = +, hece k = (+) = ( 2 ) + ( 2 3 ) ( k k+ ) = k+ s k. Such series of the form (v v + ) re clled telescopig series. I this cse, k= (v v + ) = v v k+.

28 24 Chpter 2. Numericl Series 2.2 Series with oegtive terms Lemm 2.7. Let u be series with u R +. The u coverges iff (S ) is bouded. Proof. If u for ll, the S is icresig sequece of rel umbers, which coverges iff it is bouded. Theorem 2.8 (Compriso Test). Let u d v be two series with oegtive terms. () Suppose u = O(v ). The the covergece of v implies the covergece of u, d the divergece of u implies the divergece of v. (2) Suppose u = o(v ). The the covergece of v implies the covergece of u, d the divergece of u implies the divergece of v. (3) Suppose u v. The u coverges iff v coverges. Proof. If u = O(v ), the there exists N d C > such tht u C v for ll. Let S (u) = k= u d S (v) = k= v. Suppose tht v coverges d let S v = = v. The for >, we hve S (u) = S (u) + k= + u S (u) + C k= + v S (u) + C S v. The RHS does ot deped o. Hece, (S (u)) is bouded d u coverges by Lemm 2.7. If u = O(v ) d u does ot coverge, the v does ot coverge by cotrpositio. We thus proved (). Result (2) follows becuse u = o(v ) implies u = O(v ). Result (3) follows becuse u v implies u = O(v ) d v = O(u ). Exmple If u d v re two coverget series with oegtive terms, the (mx(u, v )) coverges. Ideed, mx(u, v ) u + v. 2. Uder the sme hypotheses, if α, β re such tht α + β =, the u α v β coverges. Ideed, u α v β mx(u, v ) α mx(u, v ) β = mx(u, v ). To get more cocrete pplictios for the compriso test, we first eed to hve good referece series with which to compre. A fudmetl oe is the followig. Theorem 2. (Riem s Rule). Let α R. The α coverges iff α >. Proof. We lredy sw tht diverges i the couter-exmple of Lemm 2.4. Suppose α. The α α ( ). Ideed, α (+) α α ( α α ) ( + ) α = α = α = + + ( ( ) ) α = + α ( ( α + + o( + ))) + o(). ( ( ) ) α + But ( α (+) α ) is telescopig series, with k = ( α (+) α ) = (k+) α, which coverges iff α >. Hece, α coverges iff α >.

29 2.2. Series with oegtive terms 25 Here re two other proofs of the divergece of ) l(+) l() l( + ) l(). Ideed, / = l(+/) / = /+o(/) / = + o(). But ( l( + ) l() ) is telescopig series with k ( ) = l( + ) l() = l(k + ) l() s k, hece diverges. 2) Let S k = k =. The S 2k S k = k k. But j 2k for ech k + j 2k. Hece, S 2k S k 2k k = k 2k = 2. But if S k S, the S 2k S becuse (S 2k ) is subsequece of S k, so tht S 2k S k, cotrdictio. Hece, diverges. We ow mke importt remrk : Lemm 2.. Let u be series with oegtive terms. (i) If u for ifiitely my, the u diverges. (ii) If there exists d λ < such tht u λ for ll, the u coverges. Proof. I cse (i), u, so u diverges by Lemm 2.4. I cse (ii), u λ for ll, i.e. u = O(λ ). But λ is geometric series, which coverges becuse λ <, so u coverges. Corollry 2.2 (Root test). Let u be series with oegtive terms. Suppose tht u µ. The u coverges if µ <, d u diverges if µ >. Proof. If µ <, the for ε = µ 2, we my fid such tht u µ ε, hece u µ + ε = +µ 2 <, so u coverges by Lemm 2.. If µ >, the N with u, so u d u diverges. Note tht the root test fils whe µ =. For exmple, = d = 2 ( ) 2, but diverges while coverges. 2 ) 2 ( + 2. Exmple 2.3. Study the covergece of u, where u = ) Solutio : l u = 2 l ( + 2 thus u = e l u e 2 <. Hece, u coverges. Here is other criterio : : 2 ( 2 ) = 2. Hece, l u 2 d Lemm 2.4. Suppose there exists such tht for ll, we hve u >, v > d u + u v + v. The the covergece of v implies the covergece of u, d the divergece of u implies the divergece of v. Proof. For we hve u + v + u v, hece ( u v ) is decresig. I prticulr, u v u v d u u v v. Thus, u = O(v ) d the result follows from the compriso test. Corollry 2.5 (Rtio test). Let u be series with strictly positive terms. Suppose tht u + u λ. The u coverges if λ <, d u diverges if λ >. Proof. If λ <, the for ε = λ 2, we my fid such tht u + u λ ε, hece u + u λ + ε = +λ 2 <. Let µ = +λ 2 d v = µ. The v coverges d v + v = µ. Hece, u + u v + v d u coverges by Lemm 2.4. If λ >, the for ε = λ 2, we my fid such tht u + u λ ε, hece u + u λ ε = +λ 2 >. I prticulr, u u for ll. But u =, hece u diverges by the compriso test.

30 26 Chpter 2. Numericl Series Agi, the rtio test fils if λ =. Exmple 2.6. Study the covergece of u, where u = (2)!( k= ( + k) 2), R. Solutio : If = m for some m N, the u = for ll m, hece u coverges d equls m = u. Otherwise, u + u = (++)2 (2+)(2+2) 4, hece u is lso coverget. Theorem 2.7 (Itegrl test). Let f : [, [ R + be cotiuous d decresig. Put u = f(). The u coverges iff the sequece ( f(t) dt) coverges. Proof. Sice f decreses, we hve k+ k for ll k 2. Hece, + k+ k f(t) dt = f(t) dt f(k) f() + k= k k= k=2 k f(t) dt f(k) for ll k, d k k f(t) dt f(k) f(t) dt = f() + f(t) dt. Hece, if f(t) dt coverges, S = k= f(k) is bouded, hece f() coverges. If f(t) dt diverges, the + f(t) dt, hece S d f() diverges. Remrks I the previous proof, we estblished tht + f(t) dt k= f(k) f() + f(t) dt. Here is pplictio : study the covergece of = k= k. Solutio : let f(x) = x. The + t dt = 2 t + k= + = + (2 t k k ). Thus, 2( + ) u +2( ) d u 2 by the Sdwich theorem. 2. I cse of covergece, if we sum isted over k + i the previous proof, we obti f(t) dt f(k) f(t) dt. + k=+ Here is pplictio : estimte the remider of the series whe α >. α Solutio : let f(x) = x d R α = k=+. The α Thus, α R (+) α α 2.3 Geerl series α x α R + α x α, d R α α. α Theorem 2.9 (Cuchy criterio for series). A series of umbers u coverges iff it stisfies the Cuchy criterio : ε > N N : N, p = S +p S = u u +p ε..

31 2.3. Geerl series 27 Proof. This follows immeditely from the Cuchy criterio for sequeces, sice u coverges iff (S ) coverges. Defiitio 2.2. A series of umbers u is bsolutely coverget if u coverges. Theorem 2.2. If u is bsolutely coverget, the u is coverget. Proof. Suppose u coverges d let ε >. By the Cuchy criterio, we my fid N N such tht +p k=+ u k ε for ll N, p. Hece, +p k=+ u k +p k=+ u k ε. Hece, u coverges by the Cuchy criterio. Exmple si( 3) 3 is bsolutely coverget. If u = ( ) ( + + ), the u is ot bsolutely coverget, sice u 2. However, u is coverget. Ideed, u = v v +, where v = ( ). Hece, k= u k = k= (v k v k+ ) = v v + = + ( ) +. Remrk The compriso test of Sectio 2.2 fils whe the referece sequece chges sig. For exmple, if u = ( ) ( + + ), the = o(u ). However, u coverges, but diverges. Defiitio A rel series u is ltertig if ( ) u hs fixed sig. For exmple, ( ) is ltertig, sice ( ) ( ) = is lwys positive. Theorem 2.25 (Atertig series criterio). Suppose u = ( ) v is ltertig series such tht (v ) is oegtive, decresig d coverget to. The u coverges, d its sum S stisfies S 2+ S S 2 for ll. Moreover, R = u k v +. k=+ Proof. We prove tht S 2+ d S 2 re djcet. We hve S 2+ = S 2 + u 2 + u 2+ = S 2 + v 2 v 2+ S 2, S 2+2 = S 2 + u 2+ + u 2+2 = S 2 v 2+ + v 2+2 S 2. Thus, (S 2+ ) is icresig, (S 2 ) is decresig d S 2 S 2+ = u 2+ = v 2+. Thus, (S 2+ ) d (S 2 ) coverge to commo limit S stisfyig S 2+ S S 2 for ll. Hece, S 2 S S 2 S 2+ = v 2+, d S S 2 S 2 S 2 = v 2, which yields R 2 v 2+ d R 2 v 2. Remrk If u = ( ) v with (v ) oegtive, decresig d covergig to, the u is lso coverget, but ow S 2 S S 2+. The estimte o R is still vlid. Exmple If α >, the ( ) α I prticulr, ( ) coverges. If u = ( ) +( ), the u ( ) coverges d k=+ ( ) k k α (+) α.. However, u diverges. Ideed, u = ( ) ( ) = ( ) ( ( ) + O( ) + ( ) ) = ( ) + O( ).

32 28 Chpter 2. Numericl Series The first term gives coverget series, the third oe lso yields coverget series by Riem s rule. Hece u diverges due to the divergece of. This exmple shows gi tht the compriso test fils whe the referece sequece ) chges its sig. (here ( ) Remrk The coditio tht (v ) is decresig i Theorem 2.25 is ecessry. For exmple, if u 2 = 3 d u 2+ = +5, the u = ( ) v with (v ) positive d coverget to. However, u diverges, sice S 2+ = k= 3 k k= k Summtio by pckets Theorem Let u be series of umbers d ϕ : N N be strictly icresig. Put v = ϕ() k= u k d v = ϕ() k=ϕ( )+ u k for. The () If u coverges, the v coverges d hs the sme sum. (2) If v coverges, the u coverges i the followig cses : () u d (ϕ( + ) ϕ()) is bouded, (b) (u ) is rel, d the u k of ech pcket {ϕ( ) + k ϕ()} hve the sme sig. Proof. Let U = j= u j d V = k= v k. The V = ϕ() k= u k + ϕ() k=ϕ()+ u k ϕ() k=ϕ( )+ u k = ϕ() k= u k = U ϕ(). Thus, (V ) is subsequece of (U ). So if U u, the V coverges to the sme limit. This proves () Now suppose tht v coverges. Give N, let p be the uique turl umber stisfyig ϕ(p ) < ϕ(p). The V p U = U ϕ(p) U = ϕ(p) k=+ To prove (2.), ssume S = v d let K such tht ϕ( + ) ϕ() K for ll. Now let ε > d N N such tht u ε 2K for N. The for N we hve V p U ϕ(p) k=+ u k (ϕ(p) ) ε 2K ε 2. But there exists N such tht S V q ε 2 for q N. Give mx(n, ϕ(n )), we hve ϕ(p) ϕ(n ), so p N d thus u k. S U S V p + V p U ε, so u coverges. To prove (2.b), ote tht if ll u k hve the sme sig i ech pcket, the ϕ(p) ϕ(p) ϕ(p) V p U = u k u k = u k = v p. k=+ k=ϕ(p )+ k=ϕ(p )+ Sice v coverges, we my fid N such tht v q ε 2 for q N. Hece, for ϕ(n), we hve ϕ(p) ϕ(n), so p N d thus V p U v p ε 2. It ow follows s before tht u coverges.

33 2.5. Rerrgemets 29 Remrk 2.3. If u = ( ) d we tke pckets of size 2, i.e. ϕ() = 2 +, the v = for ll, hece v coverges but u diverges. So dditiol coditio like (2.) or (2.b) bove is ecessry to gurtee the covergece of u. Exmple 2.3. Let u = ( ) +( ). The u is ot decresig, so we cot pply the criterio of ltertig series. However, if we tke ϕ() = 2 +, the v = u 2 + u 2+ = = 2(2 + ) 4 2, d sice 2 coverges, the v coverges. Sice (2.) is stisfied, u is lso coverget. Exmple Let u = cos(2π/3). We tke pckets of size 3, i.e. ϕ() = The otig tht cos(2π/3) = cos(4π/3) = /2, we get v = u 3 + u 3+ + u 3+2 = 3 2(3 + ) 2(3 + 2) 6 2 s c be esily checked. Agi, (2.) is stisfied, so u coverges. 2.5 Rerrgemets Theorem Suppose u is bsolutely coverget d let σ : N N be permuttio, i.e. bijective mp. The u σ() is lso coverget, d u σ() = u. Proof. Let ε >. By the Cuchy criterio, we my fid N such tht, p = u u +p ε. Choose such tht the terms u,..., u pper i the prtil sum S = = u σ(). The S cotis t lest + terms, so. Hece, for m, the terms u,..., u dispper from the differece S m S m. So there exists k N such tht S m S m u u +k ε. Hece, S m S m, so S m coverges to the sme limit of S m. Remrk Absolute covergece is ecessry for the previous theorem to hold. For exmple, cosider the series u where u is give by,, 2, 2, 3, 3, 4, 4,... The tkig pckets of size 2, we see tht u coverges sice u. However, cosider the followig rerrgemet :, 2,, 3, 4, 2, 5, 6, 7, 8, 3,... 2 p +, 2 p + 2, 2 p + 3,..., 2 p, p,... The this series diverges sice it does ot stisfy the Cuchy criterio. Ideed, 2 p p p p (2p 2 p ) 2 p = 2.