1 Lecture 2: Sequence, Series and power series (8/14/2012)

Transcription

1 Summer Jump-Start Program for Aalysis, 202 Sog-Yig Li Lecture 2: Sequece, Series ad power series (8/4/202). More o sequeces Example.. Let {x } ad {y } be two bouded sequeces. Show lim sup (x + y ) lim sup x + lim sup y. Proof. Sice x + y = sup{x m + y m : m } sup{x m : m } + sup{y m : m } = x + y we have lim x + y lim (x + y ) = lim x + lim y. Therefore, lim sup (x + y ) lim sup x + lim sup y. Defiitio.. Give a sequece {x }. We say that {x j } j= is a subsequece of {x } if < 2 <... Example.2. If we cosider x = ( ), =, 2,..., the (i) {x 2 } with x 2 = is a subsequece of {x }. (ii) {x 2+ } with x 2+ = is aother subsequece. Although lim x does ot exist, lim x 2 =, ad lim x 2+ =. Theorem.. Let {x } be a sequece i R, the lim x = x for ay subsequece {x k } k=, oe has lim k x k = x. Proof. It is straight forward.

2 Propositio.. Give bouded sequece {x } i R, there exists a subsequece {x k } k= s.t. lim sup x = lim k x k. Proof. The idea for the proof: lim sup x = lim x where x = sup{x : m }. The x k /k is ot a upper boud, oe ca choose k s.t. x k x k > x k /k, iductively, we ca choose < 2 <. The lim k x k = lim sup x. Theorem.2. Bolzao-Weierstrass Theorem: Every bouded sequece i R has a coverget subsequece Proof. We cosider two cases: () {x } is fiite: there must be repeat may times. The, there is a subsequece {x k } = x. (2) {x } is ifiite. The, we choose m, M such that {x } I 0 := [m, M], we choose a half of the iterval from I k ad called I k+ with I k+ cotais ifiite set of {x }. Choose {x k } k= with x k I k. The x k x l 2M 2 k, l > k. So, {x k } is a Cauchy sequece, which has a limit. Theorem.3. Q is dese i R Because of the followig theorem o the structure of real umbers: Theorem.4. Every x R ca be represeted as x = p + j= a j 0 j, a j {0,,..., q}, p Z. So, we the get = =.0..2 Defiitio ad basic test for a coverget series Defiitio.2. We say that a coverges (to a fiite umber R), if lim S exists, where S = k= a k. Questio: How to test a coverges or diverges? Propositio.2. (Necessary coditio) If a coverges, the a 0 as. 2

3 Proof. Notice that lim a + = lim S + S = S S = 0. The proof is complete. Example.3. Prove ( ) Proof. Sice lim ( ) diverges. = 0, we have that ( ) diverges. Defiitio.3. We say that a coverges absolutely if a coverges. Notice the a {S } coverges. Oe easily has Theorem.5. (Cauchy criteria) The series a coverges if ad oly if for ɛ > 0, there is a N such that if m > N the m k=+ a k < ɛ. Corollary.6. a coverges absolutely, the a coverges. The coverse is ot true Note: First part is from: Cauchy test ad m k= a k m k=+ a k. The secod part is from ( ) coverges, but it does ot coverges absolutely..3 Series with o-egative geeral terms: a 0 Theorem.7. (Compariso test) Assume 0 b a c, =, 2,..., the (i) If c coverges, the a coverges. (ii) If b diverges (+), the a diverges. Theorem.8. p-series coverges if p > ad diverges to + if p. p 3

4 Proof. If p >, the p + + x p dx = + x p dx = + p < If p, the p + x p dx = + dx =. xp Example.4. Determie if ( ) coverges. Solutio: Sice Sice 2 coverges. 0 ( ) < + (it is a p-series, with p = 2), the by the compariso test, the series Theorem.9. (Itegral test) Assume that f(x) is a decreasig o-egative fuctio o [, ). (i) 0 a f() ad f(x)dx < +, the a coverges; (ii) a f() ad f(x)dx = +, the a diverges. Proof. (i) Sice a + f( + ) + f(x)dx, oe has a a + + f(x)dx = a + f(x)dx < +. Therefore, a coverges. (ii) Sice a f(), oe has a + a diverges. f(x)dx = f(x)dx = +. Therefore, Example.5. coverges if p > ad diverges if p. (l( + )) p Solutio. Let f p (x) = x(log(x+)) p. If p, the f(x)dx = p + l(x + ))p x= 4

5 This is fiite whe p >, ifiity whe p. Moreover, a = f(). The result follows from itegral test. Theorem.0. Ratio test for series with geeral term positive. Let a > 0, =, 2,.. ad lim a + a = r. (i) If r <, the a coverges; (ii) If r >, he a diverges; (iii) Whe r =, the test fails. Example.6. Determie if 2! coverges. Solutio: a = 2! > 0, a + = 2+! a ( + )! 2 = 2 + So, the series coverges. 0 as. Example.7. Let a = / ad b = / 2. The a + lim = b + =. a b But, a diverges ad b coverges. Theorem.. Root test If lim sup a = r, the:. if r <, the a coverges. 2. if r >, a diverges. 3. r =, test fails. Example.8. Fid iterval x such that 2 Solutio. Let a = 2 8 x3. The a = 2 8 x 3 = x x3 coverges. 2 x 3 8 <, whe x < 2. 5

6 Therefore, the series coverges for all x ( 2, 2). We have discussed how to test if a series with o-egative terms coverges. I geeral, testig the covergece of a with geeral a is very difficult. The followig idetity is a very useful tool. Theorem.2. Abel s Idetity q k=p a kb k = q k=p s k(b k b k+ ) + s q b q s p b p where s k = k l= a l, s 0 =. Proof. Notice that q q q q a k b k = (s k s k )b k = (s k b k s k b b+ ) + s q b q s p b p = s k (b k b k+ ) + s q b q s p b p. k=p k=p k=p k=p The proof is complete. Theorem.3. Test for alteratig series Let {b k } k= is a decreasig sequece ad lim b = 0. The ( ) b coverges. Proof. Apply Abel s idetity with a = ( ), s = 0 if is eve, if is odd. The the above theorem follows directly from the followig more geeral theorem Theorem.4. Let {a k } k= ad {b k} k= be two sequeces of umbers such that (i) {s = k= a k } is bouded; (ii) b b + for ad b 0 as. The k= a k b k coverges. Proof. For ay ɛ > 0, eed to fid N such that if m, > N the m k= a k b k < ɛ. Sice {s } 6

7 is bouded, there is M > 0 such that s M for all, we have m m a k b k = s k (b k b k+ ) + s m b m s b k= k= m s k (b k b k+ ) + ( s m + s )( b m + b ) k= M(b b m+ ) + 2M( b + b m ) 3M( b + b m ). Sice lim b = 0, there exists N such that if N, we have b ɛ/6m. Therefore, whe m > N, we have m k= a k b k 6M ɛ/6m = ɛ. Therefore, k= a k b k coverges by the Cauchy criteria of covergece. Example.9. Determie if ( ) l( + ) coverges. Solutio. Sice l( + 2) l( + ), =, 2,..., the lim l( + ) = 0. By the alteratig series test, Exercise: Show that ( ) coverges. l( + 2), =, 2,... ad l( + ) ( ) l( + ) coverges..4 Exercise. Prove the followig series coverges. (a) =2 (l ) 2 (b) ( ) 3 2. Prove Q Q is dese i R Ivestigate the behavior (covergece ad divergece) of a if (a) a = + ; (b) a = ( + )/; (c) a = ( / ) ad (d) a = +z, where z is complex umber. 4. If a 0 ad a coverges the a coverges. 5. If a coverges ad i {b } is mootoic ad bouded, prove that a b coverges. 6. Suppose that a > 0 ad a =. 7

8 (a) Prove that a +a diverges; (b) What ca be said about a + a, a + 2 a, (c) Let s = k= a k. Prove a s 2 s s ad deduce that a s 2 coverges. 2 Lecture 2: Sequece, Series ad power series (8/4/202) 2. More o sequeces Example 2.. Let {x } ad {y } be two bouded sequeces. Show lim sup (x + y ) lim sup x + lim sup y. Proof. Sice x + y = sup{x m + y m : m } sup{x m : m } + sup{y m : m } = x + y we have lim x + y lim (x + y ) = lim x + lim y. Therefore, lim sup (x + y ) lim sup x + lim sup y. Defiitio 2.. Give a sequece {x }. We say that {x j } j= is a subsequece of {x } if < 2 <... Example 2.2. If we cosider x = ( ), =, 2,..., the (i) {x 2 } with x 2 = is a subsequece of {x }. 8

9 (ii) {x 2+ } with x 2+ = is aother subsequece. Although lim x does ot exist, lim x 2 =, ad lim x 2+ =. Theorem 2.. Let {x } be a sequece i R, the lim x = x for ay subsequece {x k } k=, oe has lim k x k = x. Proof. It is straight forward. Propositio 2.. Give bouded sequece {x } i R, there exists a subsequece {x k } k= s.t. lim sup x = lim k x k. Proof. The idea for the proof: lim sup x = lim x where x = sup{x : m }. The x k /k is ot a upper boud, oe ca choose k s.t. x k x k > x k /k, iductively, we ca choose < 2 <. The lim k x k = lim sup x. Theorem 2.2. Bolzao-Weierstrass Theorem: Every bouded sequece i R has a coverget subsequece Proof. We cosider two cases: () {x } is fiite: there must be repeat may times. The, there is a subsequece {x k } = x. (2) {x } is ifiite. The, we choose m, M such that {x } I 0 := [m, M], we choose a half of the iterval from I k ad called I k+ with I k+ cotais ifiite set of {x }. Choose {x k } k= with x k I k. The x k x l 2M 2 k, l > k. So, {x k } is a Cauchy sequece, which has a limit. Theorem 2.3. Q is dese i R Because of the followig theorem o the structure of real umbers: Theorem 2.4. Every x R ca be represeted as x = p + j= a j 0 j, a j {0,,..., q}, p Z. So, we the get = =.0. 9

10 2.2 Defiitio ad basic test for a coverget series Defiitio 2.2. We say that a coverges (to a fiite umber R), if lim S exists, where S = k= a k. Questio: How to test a coverges or diverges? Propositio 2.2. (Necessary coditio) If a coverges, the a 0 as. Proof. Notice that lim a + = lim S + S = S S = 0. The proof is complete. Example 2.3. Prove ( ) Proof. Sice lim ( ) diverges. = 0, we have that ( ) diverges. Defiitio 2.3. We say that a coverges absolutely if a coverges. Notice the a {S } coverges. Oe easily has Theorem 2.5. (Cauchy criteria) The series a coverges if ad oly if for ɛ > 0, there is a N such that if m > N the m k=+ a k < ɛ. Corollary 2.6. a coverges absolutely, the a coverges. The coverse is ot true Note: First part is from: Cauchy test ad m k= a k m k=+ a k. The secod part is from ( ) coverges, but it does ot coverges absolutely. 0

11 2.3 Series with o-egative geeral terms: a 0 Theorem 2.7. (Compariso test) Assume 0 b a c, =, 2,..., the (i) If c coverges, the a coverges. (ii) If b diverges (+), the a diverges. Theorem 2.8. p-series coverges if p > ad diverges to + if p. p Proof. If p >, the p + + x p dx = + x p dx = + p < If p, the p + x p dx = + dx =. xp Example 2.4. Determie if ( ) coverges. Solutio: Sice 0 ( ) Sice < + (it is a p-series, with p = 2), the by the compariso test, the series coverges. 2 Theorem 2.9. (Itegral test) Assume that f(x) is a decreasig o-egative fuctio o [, ). (i) 0 a f() ad f(x)dx < +, the a coverges; (ii) a f() ad f(x)dx = +, the a diverges. Proof. (i) Sice a + f( + ) + f(x)dx, oe has a a + + f(x)dx = a + f(x)dx < +. Therefore, a coverges.

12 (ii) Sice a f(), oe has a + f(x)dx = f(x)dx = +. Therefore, a diverges. Example 2.5. coverges if p > ad diverges if p. (l( + )) p Solutio. Let f p (x) = x(log(x+)) p. If p, the f(x)dx = p + l(x + ))p x= This is fiite whe p >, ifiity whe p. Moreover, a = f(). The result follows from itegral test. Theorem 2.0. Ratio test for series with geeral term positive. Let a > 0, =, 2,.. ad lim a + a = r. (i) If r <, the a coverges; (ii) If r >, he a diverges; (iii) Whe r =, the test fails. Example 2.6. Determie if 2! coverges. Solutio: a = 2! > 0, a + = 2+! a ( + )! 2 = 2 + So, the series coverges. 0 as. Example 2.7. Let a = / ad b = / 2. The a + lim = b + =. a b But, a diverges ad b coverges. Theorem 2.. Root test If lim sup a = r, the:. if r <, the a coverges. 2

13 2. if r >, a diverges. 3. r =, test fails. Example 2.8. Fid iterval x such that 2 Solutio. Let a = 2 8 x3. The a = 2 8 x 3 = x x3 coverges. 2 x 3 8 <, whe x < 2. Therefore, the series coverges for all x ( 2, 2). We have discussed how to test if a series with o-egative terms coverges. I geeral, testig the covergece of a with geeral a is very difficult. The followig idetity is a very useful tool. Theorem 2.2. Abel s Idetity q k=p a kb k = q k=p s k(b k b k+ ) + s q b q s p b p where s k = k l= a l, s 0 =. Proof. Notice that q q q q a k b k = (s k s k )b k = (s k b k s k b b+ ) + s q b q s p b p = s k (b k b k+ ) + s q b q s p b p. k=p k=p k=p k=p The proof is complete. Theorem 2.3. Test for alteratig series Let {b k } k= is a decreasig sequece ad lim b = 0. The ( ) b coverges. Proof. Apply Abel s idetity with a = ( ), s = 0 if is eve, if is odd. The the above theorem follows directly from the followig more geeral theorem Theorem 2.4. Let {a k } k= ad {b k} k= be two sequeces of umbers such that (i) {s = k= a k } is bouded; (ii) b b + for ad b 0 as. 3

14 The k= a k b k coverges. Proof. For ay ɛ > 0, eed to fid N such that if m, > N the m k= a k b k < ɛ. Sice {s } is bouded, there is M > 0 such that s M for all, we have m m a k b k = s k (b k b k+ ) + s m b m s b k= k= m s k (b k b k+ ) + ( s m + s )( b m + b ) k= M(b b m+ ) + 2M( b + b m ) 3M( b + b m ). Sice lim b = 0, there exists N such that if N, we have b ɛ/6m. Therefore, whe m > N, we have m k= a k b k 6M ɛ/6m = ɛ. Therefore, k= a k b k coverges by the Cauchy criteria of covergece. Example 2.9. Determie if ( ) l( + ) coverges. Solutio. Sice l( + 2) l( + ), =, 2,..., the l( + 2), =, 2,... ad l( + ) lim l( + ) = 0. By the alteratig series test, ( ) l( + ) coverges. 2.4 Exercise. Prove the followig series coverge. (a) =2 (l ) 2 (b) ( ) 3 2. Prove Q Q is dese i R Ivestigate the behavior (covergece ad divergece) of a if (a) a = + ; (b) a = ( + )/; (c) a = ( / ) ad (d) a = +z, where z is complex umber. 4. If a 0 ad a coverges the a coverges. 5. If a coverges ad i {b } is mootoic ad bouded, prove that a b coverges. 4

15 6. Suppose that a > 0 ad a =. (a) Prove that (b) What ca be said about a +a diverges; a + a, a + 2 a, (c) Let s = k= a k. Prove a s 2 s s ad deduce that a s 2 coverges. 5