M17 MAT25-21 HOMEWORK 5 SOLUTIONS

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1 M17 MAT5-1 HOMEWORK 5 SOLUTIONS 1. To Had I Cauchy Codesatio Test. Exercise 1: Applicatio of the Cauchy Codesatio Test Use the Cauchy Codesatio Test to prove that 1 diverges. Solutio 1. Give the series 1, we cosider the associated series =0 1 = =0 1. Sice (1/) is a decreasig sequece which satisfies 1/ 0 for all N, ad the series =0 1 diverges, the Cauchy Codesatio test tells us that 1 diverges as well. Cauchy Sequeces. Exercise : (Abbott Exercise.6.) Give a example of each of the followig or prove that such a request is impossible. (a) A Cauchy sequece that is ot mootoe. (b) A Cauchy sequece with a ubouded subsequece. (c) A diverget mootoe sequece with a Cauchy subsequece. (d) A ubouded sequece cotaiig a subsequece that is Cauchy. Solutio. (a) This is possible. Cosider the sequece ( ) ( 1) +1 = This sequece is ot mootoe sice the eve terms are all positive ad the odd terms are all egative, however it does coverge to 0, ad is therefore Cauchy. (b) This is ot possible. The Cauchy Criterio states that all Cauchy sequeces are coverget, ad we have show i class that all subsequeces of coverget sequeces are coverget, ad that all coverget sequeces are bouded. Hece ay subsequece of a Cauchy sequece must be bouded. (c) This is ot possible. Let (a ) be a diverget mootoe sequece. The (a ) is ot bouded, sice if it were the Mootoe Covergece Theorem would imply that (a ) would coverge. Sice (a ) is ubouded ad mootoe, ay subsequece is also ubouded. If (a ) had a Cauchy subsequece, by the Cauchy Criterio, this subsequece would coverge ad would therefore be bouded, a cotradictio. Therefore (a ) cotais o Cauchy subsequeces as desired. (d) This is also possible. Cosider the sequece (a ) = (1,, 1, 3, 1, 4, 1, 5, 1, 6,...). This is clearly ubouded, however the subsequece (a 1, a 3, a 5, a 7,...) = (1, 1, 1, 1,...) is coverget ad therefore Cauchy. 1

2 M17 MAT5-1 HOMEWORK 5 SOLUTIONS Exercise 3: (Abbott Exercise.6.4) Let (a ) ad (b ) be Cauchy sequeces. Decide whether each of the followig sequeces is a Cauchy sequece. If it is, prove it. If ot, fid a couterexample. (a) c = a b. (b) c = ( 1) a. (c) c = [[a ]], where [[x]] refers to the greatest iteger less tha or equal to x. Solutio 3. (a) This is a Cauchy sequece. If (a ) ad (b ) are Cauchy sequeces, the by the Cauchy Criterio, they are coverget. Hece by the Algebraic Limit Theorem, (a b ) coverges. We showed o a previous homework that if (d ) coverges, the so does ( d ), hece ( a b ) coverges, ad is therefore Cauchy. (b) This is ot a Cauchy sequece. For example, let (a ) = (1, 1, 1,...). This sequece coverges ad is therefore Cauchy. However we have that (( 1) a ) = ( 1, 1, 1, 1,...) which does ot coverge, ad therefore by the Cauchy Criterio is ot Cauchy. (c) This is ot Cauchy. Let (a ) = (( 1) +1 /). The we have (a ) 0. However ([[a ]]) = (0, 1, 0, 1,...) which does ot coverge, ad therefore by the Cauchy Criterio is ot Cauchy. Cauchy Series. Exercise 4: Cauchy Series Show that a series k=1 a k is Cauchy if ad oly if for all ɛ > 0 there exists N N such that wheever > m N, we have a m a < ɛ. Solutio 4. ( ). Suppose that k=1 a k is a Cauchy series, ad ote that by defiitio this meas that the sequece of partial sums (s m ) is a Cauchy sequece. Let ɛ > 0 be arbitrary. Sice (s m ) is a Cauchy sequece, there exists N N such that if, m N, we have s s m < ɛ. We may assume without loss of geerality that > m, ad the we compute a m a = s s m < ɛ, as desired. ( ). Let ɛ > 0 be arbitrary. By hypothesis, there exists N N such that if > m N, the a m a < ɛ. Therefore if, m N, the without loss of geerality, > m, ad we have s s m = a m a < ɛ, so (s m ) is a Cauchy sequece ad hece k=1 a k is a Cauchy series.

3 Algebraic ad Order Limit Theorems for Series. M17 MAT5-1 HOMEWORK 5 SOLUTIONS 3 Exercise 5: Algebraic Limit Theorem for Series (Abbott Theorem.7.1) Complete the proof of the ALT for Series: Show that if k=1 a k = A ad k=1 b k = B, the the followig cosequece holds. (b) k=1 a k + b k = A + B. Solutio 5. (b) Note that if (s m ) deotes the sequece of partial sums of k=1 a k ad (t m ) deotes the sequece of partial sums of k=1 b k, the for every m N, we have s m + t m = (a a m ) + (b b m ) = (a 1 + b 1 ) (a m + b m ) Therefore (s m + t m ) is the sequece of partial sums of k=1 a k + b k. By the Algebraic Limit Theorem for sequeces, sice (s m ) A ad (t m ) B, it follows that (s m + t m ) A + B, therefore by the defiitio of covergece of a series, we have that k=1 a k + b k = A + B. Exercise 6: (Abbott Exercise.7.4) Give a example of each of the followig or prove that such a request is impossible. (a) Two series x ad y that both diverge, but where x y coverges. (b) A coverget series x ad a bouded sequece (y ) such that x y diverges. (c) Two sequeces (x ) ad (y ) where x ad (x + y ) both coverge but y diverges. (d) A sequece (x ) satisfyig 0 x 1/ where ( 1) x diverges. Solutio 6. (a) Let (x ) = (( 1) +1 /) = (1, 1, 1, 1,...) ad let (y ) = (1/) = (1, 1/, 1/3, 1/4,...). The x diverges because its sequece of terms does ot coverge to 0, ad y is the harmoic series, which diverges. However x y = which is the alteratig harmoic series, which we have ( 1) +1 show doverges. (b) This is ot possible. Let (x ) = (y ) = (( 1) +1 / ). The Alteratig Series Test shows that x coverges, ad it is clear that (y ) coverges to 0, ad therefore (y ) is bouded. Yet we have a b = 1/, which diverges. (c) This is ot possible. If x ad (x +y ) coverge, the the algebraic limit theorem states that x coverges, ad that y = ((x + y ) + ( x )) coverges. (d) Cosider the sequece (x ) = (0, 1/, 0, 1/4, 0, 1/6,...), which clearly satisfies 0 x 1/. Note that (x ) = (0, 1, 0, 1/, 0, 1/3, 0, 1/4,...), so the series x diverges, which implies, by the Algebraic Limit Theorem, that x diverges as well. Note that (( 1) x ) = (x ), ad therefore ( 1) x diverges, as desired. Exercise 7: (Abbott Exercise.7.7) (a) Show that if a > 0 ad lim (a ) = l with l 0, the the series a diverges. (b) Assume a > 0 ad lim ( a ) exists. Show that a coverges. Solutio 7.

4 M17 MAT5-1 HOMEWORK 5 SOLUTIONS 4 (a) If lim (a ) = l ad a > 0, it follows that l 0. But by hypothesis, l 0, hece l > 0. The for ɛ = l/, there exists N N such that for all N, we have a l < ɛ = l/. Hece l/ < a l < l/ which implies that l/ < a, ad hece that a > l/. Sice 1 diverges, it follows that l/ diverges as well. Sice for all N the terms of a are larger tha the terms of, it follows from the compariso test that a diverges. (b) Let lim a = l. The for ɛ = 1, there exists N N such that for all N, we have a l < 1, ad therefore that a < l + 1, which implies that 0 < a < l+1. Sice 1 coverges, so does l+1. Therefore, sice the terms of a satisfy a < l+1 for all N, it follows from the compariso test that a coverges. Absolute Covergece. Exercise 8: Coverse of the Absolute Covergece Test Does covergece of k=1 a k imply absolute covergece of k=1 a k? Prove or fid a couterexample. l/ Solutio 8. No. Recall that the Alteratig Series Test implies 1 which diverges. ( 1) +1 coverges, yet ( 1)+1 = Exercise 9: (Abbott Exercise.7.8) Cosider each of the followig propositios. Provide proofs for those that are true ad couterexamples for those that are ot. (a) If a coverges absolutely, the a also coverges absolutely. (b) If a coverges ad (b ) coverges, the a b coverges. (c) If a coverges coditioally, the a diverges. Solutio 9. (a) This is true. Sice a coverges, we have that (a ) 0. Hece there exists N N such that if N, the a < 1. This implies that for N, 0 a a. Therefore (a ) differs at oly fiitely poits from a series (x ) which satisfies 0 x a. Sice a coverges, it follows from the order limit theorem that x coverges as well. Sice it oly differs from (a ) at fiitely may poits, it follows that a coverges as well. (b) This is ot true. Let (a ) = (b ) = (( 1) +1 / ). The Alteratig Series Test shows that a coverges, ad it is clear that (b ) coverges to 0, yet a b = 1/, which diverges. (c) This is true. Suppose that a coverges to some l R. The we have that the sequece of terms ( a ) coverges to 0. Therefore there exists N N such that if N, we have a < 1. But this implies that for N, we have a < 1. Therefore (a ) oly differs at fiitely may poits from a sequece (x ) which satisfies 0 < x < 1.The Compariso Test implies that (x ) coverges. Sice ( a ) oly differs from (x ) at fiitely may terms, so a coverges, ad a coverges absolutely. Geometric Series.

5 M17 MAT5-1 HOMEWORK 5 SOLUTIONS 5 Exercise 10: (Abbott Exercise.5.7) Let b R. Prove that the sequece (b ) coverges to 0 if ad oly if b < 1. Solutio 10. ( ). Suppose that (b ) coverges to 0, ad suppose that b 1, the b > b 1, so the terms of (b ) are bouded away from 0 by 1. It follows that (b ) does ot coverge. ( ). Suppose that b < 1. If b = 0, the we have that (b ) = (0, 0, 0,...) which coverges to 0. If 0 < b < 1, the we have already show i class that (b ) 0. Now suppose that 1 < b < 0, ad ote that ( b ) 0, agai by what we showed i class. Let ɛ > 0 be arbitrary. The sice ( b ) 0, there exists N N such that if N, we have b 0 < ɛ, which implies that b < ɛ, which implies that b 0 < ɛ. Therefore we have show that for all ɛ > 0, there exists N N such that for all N, we have b 0 < ɛ. This implies that (b ) 0 i this case as well. Exercise 11: Ratio Test (Abbott Exercise.7.9) Give a series a with a 0, the Ratio Test states that if a satisfies lim a +1 a = r < 1 the the series coverges absolutely. (a) Let r satisfy r < r < 1. Prove that there exists N N such that for all N, we have a +1 < a r. (b) Prove that a N (r ) coverges. (c) Now, show that a coverges, ad coclude that a coverges absolutely. Solutio 11. (a) Let r satisfy r < r a < 1. Sice lim +1 a = r < 1, let ɛ = r r > 0. The there exists N N such that for all N, we have a +1 r < ɛ a = = = a +1 a a +1 a a +1 a = a +1 < a r, < r + ɛ < r + r r < r as desired. (b) Sice 0 r < 1, it follows that (r ) is a coverget geometric series. Sice a N is just a costat, a N (r ) coverges as well. (c) Let ɛ > 0 be arbitrary. The by part (a), there exists N N such that if N, the a +1 < a r. Therefore if N + 1, it follows that a < a 1 r. Repeated applicatio of this iequality yields a < a N (r ) N < a N (r ). Therefore by the compariso test ad part (b), it follows that a coverges, ad hece coverges absolutely. a

6 M17 MAT5-1 HOMEWORK 5 SOLUTIONS 6 Exercise 1: Root Test Give a series a with bouded terms, let C = lim sup a The root test states that: If C < 1, the the series coverges absolutely. If C > 1, the series diverges. If C = 1, the the test is icoclusive. (a) Show that if C < 1, the there exists 0 k < 1 ad there exists some N N such that for all N, we have a k. The use geometric series ad the compariso test to show that a coverges absolutely. (b) Show that if C > 1, the a > 1 for ifiitely may, ad therefore the sequece of terms (a ) does ot coverge to 0. Use this to show that a diverges. (c) Give a example of a absolutely coverget series with C = 1 ad aother example of a diverget series with C = 1. You may assume, for part (c), that lim = 1. Solutio 1. (a) Let x = sup{ k a k k } so that lim sup a = lim x. Suppose that C < 1, let ɛ = 1 C > 0, ad let k = C + ɛ = C+1 < 1. The there exists N N such that if N, x C < ɛ, which implies that x < C + ɛ = k. Therefore, i particular, we have that x N < k, which implies that 0 a < k for all N, as desired. Takig the th power of the iequalities above yields 0 a < k for N. Sice 0 k < 1, the series k is a coverget geometric series, so by the compariso test, a coverges, ad hece a coverges absolutely. (b) If C > 1, the (x ) is a decreasig sequece which coverges to C. Sice x = sup{ k a k k }, for each N, there is some k such that C k a k x. This implies that 1 < k a k for all such k, which implies that 1 < a k for ifiitely may k N. It therefore follows that (a k ) does ot coverge to 0, hece a must diverge. (c) Let a = 1. The the series 1 is the harmoic series ad therefore diverges. Note that because lim 1 = 1, it follows that lim = 1, hece lim sup 1 = 1, so C = 1. O the other had, the series 1 coverges absolutely, ad ( ) 1 1 =. ( ) 1 1 Therefore, by the algebraic limit theorem, lim = lim = 1, so lim sup 1 = 1, ad we have that C = 1.. Practice Problems (Do t had these i)