Math 312, Intro. to Real Analysis: Homework #4 Solutions

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1 Math 3, Itro. to Real Aalysis: Homework #4 Solutios Stephe G. Simpso Moday, March, 009 The assigmet cosists of Exercises 0.6, 0.8, 0.0,.,.3,.6,.0,.,. i the Ross textbook. Each problem couts 0 poits (a) Let (s ) be a sequece of real umbers. The, wheever m is less tha we have s m s = (s m s m+ ) + (s m+ s m+ ) + + (s s ), hece by the Triagle Iequality s m s s m s m+ + s m+ s m+ + + s s. Now, if our sequece is such that s k s k+ / k for all k, the our iequality becomes s m s m + m+ + + ad the right-had side is the begiig of the geometrical series k=m /k = / m. Thus we see that s m s / m for all m ad all > m. Thus (s ) is a Cauchy sequece. Therefore, by oe of our theorems, (s ) is a coverget sequece. (b) If we assume oly that s k s k+ < /k for all k, we caot coclude that (s ) is a Cauchy sequece. A example illustratig this is obtaied by lettig s = the th partial sum of the harmoic series, s = H = For this example we have but s s + = + < lims = = so our sequece (s ) is diverget. = = + (harmoic series)

2 0.8. Assume that (s ) is a odecreasig sequece of real umbers. Let σ be the average of the first umbers i our give sequece: σ = s + + s. We claim that the sequece (σ ) is agai odecreasig. To see this, ote that s s s +, hece s + + s s +. Addig (s + + s ) to both sides, we obtai (s + + s ) + (s + + s ) (s + + s ) + s +, or, i other words, ( + )(s + + s ) (s + + s + s + ). Dividig both sides by ( + ), we obtai s + + s s + + s + s + + or, i other words, σ σ + for all. This proves our claim. The coclusio ca be restated as follows: If a sequece is odecreasig, the its sequece of averages is odecreasig. Note: The textbook assumes i additio that s is positive for all. However, this assumptio is ot eeded i order to obtai the coclusio Defie a sequece (s ) iductively by lettig s = ad s + = (s +)/3 for all. (a) The first few terms of the sequece are s = ad s = /3 ad s 3 = 5/9 ad s 4 = 4/7. (b) Obviously s > /. If s > /, it follows that s + > 3/, hece s + = (s + )/3 > (3/)/3, hece s > /. Thus by iductio o we see that s > / for all. (c) Sice s > /, it follows that s s + = s s + 3 = 3 s 3 > 3 3 = 0 so s > s + for all, i.e., (s ) is a decreasig sequece. (d) Sice (s ) is a decreasig sequece which is bouded below (the boud is /), it follows by oe of our theorems that (s ) is coverget. Let α = lims. Takig the limit of both sides of the equatio s + = (s + )/3, we obtai α = (α + )/3. Solvig for α we obtai α = /. Thus lims = /.

3 .. Let a = ( ), b =, c =, d = Note that the sequece (a ) looks like,,,,.... (a) A mootoe subsequece of (a ) is,,,.... We ca describe this subsequece as (a k ) k where k = k. Thus a k = ( ) k = for all k. Alteratively, usig istead of k as the idex, we ca describe our subsequece as (a ). The sequeces (b ), (c ), ad (d ) are already mootoe. (Note that ay sequece is cosidered to be a subsequece of itself.) To see that (d ) is mootoe, use algebra to show directly that for all ( + ) + 4 > 7 3 7( + ) 3 (b) The subsequetial limits of our sequeces are ad for (a ), 0 for (b ), + for (c ), ad 6/7 for (d ). (c) limsup a =, limif a =. limsupb = limif b = limb = 0. limsup c = limif c = limc = +. limsupd = limif d = limd = 6/7. (d) (a ) is diverget ad its limit is udefied. (b ) ad (d ) are coverget to 0 ad 6/7 respectively. (c ) is diverget to +. (e) (a ), (b ), ad (d ) are bouded. (c ) is ubouded..3. Let s = cos π 3, t 3 = 4 +, ( u = ), v = ( ) +. Note that the sequece (s ) looks like,,,,,,,,,.... (a) A mootoe sbsequece of (s ) is,,,..., which ca be described as (s 6+3 ), i.e., (cos( + )π). The sequece (t ) is already mootoe. A mootoe subsequece of (u ) is (u ), i.e., (/ ), which looks like /4, /6, /64,.... A mootoe subsequece of (v ) is (v ), i.e., ( + /), which looks like 3/, 5/4, 7/6,.... (b) The subsequetial limits are ±, ±/ for (s ), 0 for (t ), 0 for (u ), ± for (v ). (c) (s ) ad (v ) are diverget ad their limits are udefied. (t ) ad (u ) are coverget to 0. (d) (s ), (t ), (u ), ad (v ) are bouded. 3

4 .6. Assume that (c ) is a subsequece of (b ) ad (b ) is a subsequece of (a ). By equatio (3) o page 64, there are fuctios f() ad g() such that b = a f() ad c = b g() for all. It follows that c = a f(g()) for all. Thus (c ) is a subsequece of (a ) via the composite fuctio f g..0. We are give the sequece, /,,, /, /3, /4, /3, /,,, /,.... The subsequetial limits of this sequece are, /, /3,..., /,... ad 0. The limsup is ad the limif is 0... By defiitio of limits, lims = 0 meas that for all ǫ > 0 we ca fid N so large that s 0 < ǫ for all > N. I other words, for all ǫ > 0 we ca fid N so large that s < ǫ for all > N. But the, by the defiitio of lim sup, this coditio is precisely equivalet to sayig that limsup s 0. Moreover, s 0 automatically, hece limsup s 0 automatically, for all sequeces. Thus, our coditio above is equivalet to sayig that limsup s = 0. We have ow proved that lims = 0 if ad oly if limsup s = 0... Assume that (s ) is a sequece of real umbers. As i Exercise 0.8, let σ be the average of s,..., s. (a) We shall prove the chai of iequalities limif s limif σ limsup σ limsup s. () The middle iequality limif σ limsup σ is obvious ad does ot require proof. We prove the right-had iequality limsup σ limsup s. We omit the proof of the left-had iequality limif s limif σ, which is similar. I order to prove limsupσ limsup s, we follow the hit give i the textbook. First, give M ad N such that M > N, we claim that sup σ s + + s N >M M To prove (), it suffices to prove σ s + + s N M + sup s k. () + sup s k (3) for each idividual > M. Sice > M > N, we ca break σ ito two parts: σ = s + + s = s + + s N + s N+ + + s. (4) Sice > M, the first part of the right-had side of (4) is < the first part of the right-had side of (3). Ad obviously, the secod part 4

5 of the right-had side of (4) is the secod part of the right-had side of (3). Combiig these two observatios, we see that (3) holds. From this it follows immediately that () holds. Now, fixig N ad takig the limit of () as M goes to ifiity, we see that limsup σ sup s k (5) for all N. Fially, takig the limit of (5) as N goes to ifiity, we see that limsup σ limsup s k ad this completes the proof. (b) I the case whe lims exists, we have lim s = limif s = limsups, so () becomes limif s = limif σ = limsupσ = limsups = lims. From this it follows that lim σ exists ad is equal to lims. This coclusio ca be restated as follows: If a sequece is coverget, the its sequece of averages is coverget. 5