Math 230a Homework 4

Transcription

1 Math 30a Homework 4 Erik Lewis December 0, 006 Rudi. Prove that the covergece of {s } implies the covergece of { s }. Is the coverse true? Proof: Suppose {s } coverges to s. So we kow that s s < ɛ. We earlier proved from the triagle iequality that a b a b. If we replace a ad b with s ad s we get s s s s. So we have that s s s s < ɛ. So we have that { s } coverges to s. But sice it is a iequality, we do ot ecessarily have that if { s } coverges the {s } coverges.. Calculate ( + ). Proof: If we multiply by the cojugate we get: ( + ) ( + + ) ( + + ) = = = = + ( + + ) ( + + ) ( + + ) ( + + ) =

2 3. If s =, ad s + = + s ( =,, 3,...), prove that {s } coverges, ad that s < for =,, 3,.... Proof: We ca see that 0 < < < + < 4 < + <. Applyig this rule agai we get that + < + + <. So we ca guess that s < + s < N. To prove this use iductio. We have the case for = : < + <. So ow, assume its true for ad show its true for +. So s + = + s < + + s < s + < + s + = s + <. So we have a fuctio that is always icreasig ad bouded above by so the sequece coverges by the Mootoe Covergece Priciple. 4. Fid the upper ad lower its of the sequece {s } defied by s = 0; s m = s m ; s m+ = + s m. Proof: If we look at the first few terms of the sequece we have that: s m = 0, 4, 3 8, 7 6,... s m+ = 0,, 3 4, 7 8, 5 6,... By ispectio we ca see that the if s = if s = ad sup s = sup s + =. 5. For ay two real sequeces {a }, {b }, prove that sup (a + b ) sup provided the sum o the right is ot of the form. Proof: We kow that a + sup b, sup{a + b, a + + b +,...} sup{a, a +,...} + sup{b, b +,...} sup{a + b, a + + b +,...} (sup{a, a +,...} + sup{b, b +,...}) sup(a + b ) (sup{a } + sup{b }) sup (a + b ) sup a + sup b

3 6. Ivestigate the behavior (covergece or divergece) of a if (a) a = + ; Proof: a = ( + ) ( ++ ) ( ++ ) = + ++ = ++. So a diverges by the p-series test. (b) a = + ; Proof: a = ( + ) ( ++ ) ( ++ = ) + ( ++ = ) ( ++ ). So a coverges by the p-series test with p = 3/. (c) a = ( ) ; Proof: Use the root test: a = ( ) = sup a = sup So sup a = 0 < a coverges. = = 0 7. Prove that the covergece of a implies the covergece of if a 0. Proof: By the Schwarz Iequality: a, N ( = N = a ) a N = = a N = = N ( N a ) ( ) ad this is true for all N N. Sice coverges by the p-series test ad we are give that a coverges, we ca say: N = a N ( N a ) ( ) = = a = = Sice both series are coverget we ca say the product of ay fiite sum of terms is less tha the ifiite sum. So we have that a is always bouded above by a <. Sice a 0 for all, we kow that a 0. Sice the terms are all positive we kow the sequece of partial sums {s } is bouded above ad mootoic icreasig, so we have that the sum coverges. 3

4 8. If a coverges, ad if {b } is mootoic ad bouded, prove that a b coverges. Proof: Cosider the case where {b } is mootoe icreasig ad bouded above by say B, the b B for all. So the a b a B = B a = AB where A is the ifiite sum of the a s. The other case is if {b } are decreasig ad bouded below. The we ca ot use the compariso test with the lower boud of {b }. But we do kow that b < b for all > because the sequece is mootoe decreasig. So we have that a b a b = b a = b A. So i either case a b coverges. 9. Fid the radius of covergece of each of the followig series: (a) 3 z Proof: Use the root test: (b)! z α = sup 3 = sup( / ) 3 = R = α =. Proof: The! is harder to evaluate so use the ratio test: α = sup + (+)!! = sup +! ( + )! = sup + = 0 R = α =. 0. Suppose that the coefficiets of the power series a z are itegers, ifiitely may of which are distict from zero. Prove that the radius of covergece is at most. Proof: Sice a z coverges we kow that a z = 0. Sice the it exists we ca say that a z = 0. By assumptio ifiitely may of the a s are itegers distict from zero so we kow that it is ot zero. This meas z = 0 z < R = =. So the radius of covergece is at most.. Suppose a > 0, s = a a, ad a diverges. (a) Prove that a +a diverges Proof: Assume to the cotrary that a +a coverges. That meas that has to go + a to zero which meas a = a = 0. From the Cauchy Criterio for covergece we kow that there exists ad N > 0 such that for all > N, a <. If that is true the we ca say that a a +a = a +a < a +a. So that meas that a a +a coverges because by assumptio a +a coverges. But that meas that a a +a + a +a = a coverges. But 4

5 this cotradicts the assumptio i the begiig of the problem that a diverges. So diverges. (b) Prove that ad deduce that a s diverges. Proof: We kow for sure that a N+ s N a N+k = s N s N + a N a N+k = a N a N+k = s N a N a N+k = s N a +a Because each a > 0, if we exclude the right terms from each of the deomiators o the left the we will make each term larger ad thus make the left had side larger. So a N+ s N a N+k s N To show that the a s all ɛ > 0 there exists a N > 0 such that for all > m > N, the there is some N > 0 such i=m that am s m a s (c) Prove that diverges, assume to the cotrary that it coverges. The we kow that for a i i=m s i < ɛ. So let ɛ = sm s > 0 a i s i < ɛ = sm s > 0. But we have from the iequality above which is a cotradictio. So it must be the case that a s diverges. sm s a s s s ad deduce that a s coverges. Proof: Start with the left had side of the iequality ad use the fact that s < s a s = s s s s s s s s = s s which is what we were tryig to prove. Now look at the the sums of both sides of this iequality: a s + k= a k s k a s + s k s k k= The right had side is a telescopig sum so we have that a s + k= a k s k a s + s s 5

6 (a s + k= a k s ) (a k s + ) s s Sice a > 0 that meas s is always mootoe icreasig. So we ca say that: a s a s + s s Sice the partial sums of a s coverges as well. a s (d) What ca be said about a ad + a a s + = s are mootoe icreasig ad bouded above by a + s s so the a s a + a? Proof: For a = the first sum diverges while for the example i the extra credit it coverges so we ca ot say aythig about its covergece or divergece from the iformatio give. As for the secod oe it coverges. To see this compare it to : Sice coverges so does a + a. a + a = a + a. Suppose a > 0 ad a coverges. Put r = a m. m= (a) Prove that a m a r r m r r m if m <, ad deduce that a r diverges. Proof: Sice a coverges, we kow that removig a fiite umber of terms at the begiig does ot chage the covergece of the series so each r coverges as well. So r m r m = a m a + r r m = a m a = r r m r m r m 6

7 Because each a > 0, we kow that r i < r m for all m < i. So chagig each deomiator from r m to r i will make the left had side larger. So we get the desired result that a m r m a r r r m To show that a r diverges, lets assume to the cotrary that it coverges. The that meas for all ɛ > 0 there exists a N > 0 such that for all > m > N a i i=m r i < ɛ. So let ɛ = r r m > 0 the there is some N > 0 such that a i i=m r i < ɛ = r r m > 0. But we kow from the iequality above that am r m a r r r m. So it must be the case that a r diverges. (b) Prove that ad deduce that a r coverges. a r < ( r r + ) Proof: Sice r = a + a ad r + = a + + a we kow that a = r r +. So r r + = a a = ( r r + )( r + r + ) a r + = r r + r + a r + r = r r + a r = r r + a r < ( r r + ) As i the problem before, lets look at the sums of both sides k= k= a r < ( r r + ) k= a r < k= ( r r + ) O the right had side we agai have a telescopig sum. So we cotiue with k= a r < ( r r + ) a r < r 7

8 So r a is mootoe icreasig ad bouded above. So it coverges by the Mootoe Covergece Priciple. 3. Prove that the Cauchy product of two absolutely coverget series coverges absolutely. Proof: Let a ad b are the two absolutely coverget series. The we kow that a ad b both coverge as well. From Merte s Theorem we kow that the Cauchy Product d of a ad b will coverge. But this is ot quite what we wat. We wat the Cauchy product c of a ad b to coverge absolutely. But we do kow that c d for all. So we have that c d c coverges absolutely by the compariso test. 4. If {s } is a complex sequece, defie its arithmetic meas σ by σ = s 0 + s s + ( = 0,,,...). (a) If s = s, prove that σ = s. Proof: If ɛ > 0, we wat to fid a N > 0 such that (s0+...+s) + l < ɛ, N. So we have that s N+ l < ɛ, s N+ l < ɛ,..., s N+M+ l < ɛ ad if we put them all together we get s N+ l + s N+ l s N+M+ l < (M + )ɛ Also, we ca say that s N+ + s N s N+M Ml < Mɛ As well we kow that ad s N+ + s N a N+M+ N + M s 0 + s + s s N < ɛ N + M + (M + )l N + M + l < ɛ (M + )l (M + )ɛ < N + M + N + M + < ɛ if we pick a appropriate N. Fially, if we put all three together we get that: s N+ + s N s N+M+ N + M + (M + )l N + M + + s 0 + s + s s N N + M + s 0 + s + s s N+M+ N + M + l < 3ɛ + (M + )l l < 3ɛ N + M + Sice 3 is just a costat, we could go back ad adjust our ɛ s but it wo t chage the proof. (b) Costruct a sequece {s } which does ot coverge, although σ = 0. 8

9 Proof: Look at the sequece s = ( ). The sequece obviously does t coverge because its sup is ot equal to its if. But σ 0 as because we get that the s 0 + s s is equal to either or -, while the deomiator goes to zero. (c) Ca it happe that s > 0 for all ad that sup s =, although σ = 0? Proof: Let a = log log( + ) the we have that σ = log log +log log +...+log log(+) + = log(log() log()... log(+)) log( log ) + +. The σ = 0 but sup s = albeit slowly. Spivak 4. (a) Prove that if a subsequece of a Cauchy sequece coverges, the so does the origial sequece. Proof: Let {x } be the Cauchy sequece, ad let {x k } be the coverget subsequece ad call its it x. Suppose ɛ > 0 the there exists a N such that d(x, x m ) < ɛ/ wheever, m > N. Sice x k is a coverget subsequece there exists 0 > N with d(x 0, x) < ɛ/. If m > the d(x m x) d(x m, x 0 ) + d(x 0, x) < ɛ/ + ɛ/ = ɛ. So x x. (b) Prove that ay subsequece of a coverget sequece coverges. Proof: Let {x } coverge to x. The for all ɛ > 0 there exists a N > 0 such that d(x, x) < ɛ for all > N. Every coverget sequece is Cauchy, so that meas there exists M such that d(x, x m ) < ɛ for all, m > M. So cosider a subsequece {x k } of the sequece. The there exists a K such that k > max{n, M} = J for all k > K. So the we have that d(x k, x) d(x k, x J ) + d(x J, x) < ɛ + ɛ = ɛ. So the subsequece {x k } coverges as well to x. 5. (a) Prove that if 0 < a <, the a < a <. Proof: We kow that 0 < a < 0 < a < a < 4 a < a <. (b) Prove that the sequece coverges.,,,... Proof: So the sequece ca be be writte as a + = a. From part (a) we kow that < <, as well we kow that < <. So we ca guess that a < a < N. To prove this use iductio. We have the case for = : < <. So ow, assume its true for ad show its true for +. So a + = a < a < a + < a + <. So we have a fuctio that is always icreasig ad bouded above by so the sequece coverges by the Mootoe Covergece Priciple. 9

10 (c)fid the it. Hit: Notice that if a = l, the a = l, by Theorem. Proof: Sice a exists from part (b), the lets say that a = l. We also kow that a = a + = a = l. So we have that l = l l l = 0 l = 0 or l =. Sice a > 0 for all we kow that a =. 6. Let 0 < a < b ad defie a + = a b, b + = a + b. (a) Prove that the sequeces {a } ad {b } each coverge. Proof: We have from the first homework that if 0 < a < b ad a < ab < a+b < b. We kow 0 < a < b so a < a b < a+b < b a < a < b < b. So we wat to show that a < a < b < b. We have that it is true for =. Assume it is true for, the we have that a < a < a b < a+b < b < b a < a < a + < b + < b < b a < a + < b + < b. So we have that each a is bouded above by ay b ad it is icreasig so a coverges by the Mootoe Covergece Priciple. The same is true for b except that it is bouded below by each a ad decreasig so b coverges by the Mootoe Covergece Priciple. (b) Prove that they have the same it. Proof: From part a) we kow both the sequeces coverge. So say the it of {a } is a, ad a the it of {b } is b. a = a b ad b = +b b = a+b b = a + b a = b. So they have the same its. 8. Idetify the fuctio f(x) = ( k (cos!πx) k ). Proof: Cosider the cases separately whe x is irratioal ad whe x is ratioal. Assume x is ratioal. The x = p/q for some p, q N. That meas f(p/q) = ( k (cos!π(p/q)) k ). If we let N = q, the > N we will have that ( k (cos!π(p/q)) k ) = because every value of!π(p/q) will be divisible by π if > N. So that meas that ( k (cos!π(p/q)) k ) = ( k () k ) =. Now cosider whe x is irratioal. That meas that for all N,!x will always be irratioal. So we have that cos(!πx) < k (cos!πx) k = 0 ( k (cos!πx) k ) = 0. So we have that f(x) = { if x is ratioal 0 if x is irratioal 9. May impressive lookig its ca be evaluated easily, because they are really upper or lower sums i disguise. With this remark as a hit evaluate each of the followig. (Warig: oe of these ca be evaluated by elemetary cosideratios.) (iv) ( + (+) () ). 0

11 Proof: Each term i the sequece is less tha. So we ca compare the it to: (v) ( (+) + ( ) = = 0 ( + ( + ) () ) = 0. (+) (+) ). Proof: We ca rewrite the part i the it as a itegral. To see this i= ( + i) = If we iterpret this as a itegral we get: i= ( + i = ) i= ( + i ) = i= x dx = (vi) ( ). Proof: Agai, we ca rewrite this it as a it of a sum: i= + i = If we agai iterpret this as a itegral we get: i= + i = 0 i= + i + x dx = arcta x 0 = π ( + i ) 0. Although its like ad a ca be evaluated usig facts about the behavior of the logarithm ad expoetial fuctios, this approach is vaguely dissatisfyig, because itegral roots ad powers ca be defied without usig the expoetial fuctio. Some of the stadard elemetary argumets for such its are outlied here; the basic tools are iequalities derived from the biomial theorem, otably ad, for part (e), ( + h) + h + ( + h) + h, forh > 0; ( ) h ( ) h, forh > 0. (a) Prove that a = if a >, by settig a = + h, where h > 0.

12 Proof: Let a = + h the a = ( + h) + h =. So we have that a a =. (b) Prove that a = 0 if 0 < a <. Proof: We kow that 0 < a < 0 < < a. So the if we let a = b, the by part a) we have that b = b = a = = 0. (c) Prove that a = if a >, by settig a = + h ad estimatig h. Proof: So followig the hit let a = + h a = ( + h) + h for all h > 0. So the we ca solve for h to get h a < + h + a < ( + h) ( + a ) < a (d) Prove that a = if 0 < a <. a = Proof: We kow that 0 < a < 0 < < a. The a = a = a =. So we have that (e) Prove that =. Proof: We ca use the hit to say that So o we have a boud for h we ca say that: = + h = ( + h) ( ) h h = + h + + Sice = + h that implies that which meas that =.. (a) Prove that a coverget sequece is always bouded.

13 Proof: Sice the sequece coverges, it is Cauchy. So that meas for ɛ = there is a N > 0 such that for all, m > N we have that d(a, a m ) <. More importatly, this is true for all, m > N so that meas d(a N+, d m ) < for all m > N. So let M = {d(x, x N+ ), d(x, x N+ ),..., d(x N, x N+ ), }. The d(x k, x N+ ) M for all k =,, 3... (b) Suppose that a = 0, ad that each a > 0. Prove that the set of all umber a actually has a maximum member. Proof: Sice the sequece coverges to zero, we ca say that there exists a N such that a 0 < ɛ = a, for all > N. The the max of a,..., a N is max of a for all because all the a s after N will be smaller tha a so we eed oly cosider the oes from to N. So a has a maximum member.. (a) Prove that ( + ) < log( + ) log <. Proof: Cosider the itergral + x dx. The the smallest lower sum is just the smallest fuctio value o the iterval [0,] evaluated at the partitio P = {, + }. This gives f( + ) ( + ) = +. Ad if we use the same partitio P ad just evaluate at the largest value of f we get. Sice the fuctio is cotiuous we kow the fuctio falls somewhere i betwee these two values, which gives us our desired result that (+) < log( + ) log <. (b) If a = log, show that the sequece {a } is decreasig, ad that each a 0. It follows that there is a umber γ = ( log ). This umber, kow as Euler s umber, has proved to be quite refractory; it is ot eve kow whether γ is ratioal. Proof: To show that the sequece is decreasig, we wat to show that a + a 0 a a +. So we have: a a + = ( log ) ( log( + )) + a a + = log + log( + ) + From part (a) we kow that (+) < log( + ) log, so we kow have that a a + = log( + ) log + > 0 3

14 a + < a. As i part (a) we ca look at the upper sum of the xdx. If we cosider the partitio P = {,, 3,..., }, the the upper sum of this partitio is just the fuctio values at right edpoits because the fuctio is mootoe decreasig ad cotiuous. The we get that f() + f(3) f() > x dx = log log = log log > log > 0 a > (a) Suppose that f is icreasig o [, ). Show that f() f( ) < f(x)dx < f() f(). Proof: Sice f(x) is icreasig o the iterval [, ] the this looks like upper ad lower sums. The left had side is k= f(x i) x i ad the right had side is k= f(x i) x i where x i =. Sice the itegral is always betwee ay upper ad lower sum for a give partitio if the fuctio S(f, P ) f(x)dx S(f, P ) Ad this is exactly what we are tryig to prove for ay give partitio of f(x) where x i =. (b) Now choose f = log ad show that it follows that ( + )+ <! < e e! = e. Proof: Sice log x is icreasig o [, ), it follows from part (a) that: log() log( ) < Lookig at just the right iequality we have Lookig at just the left iequality we have log(x)dx < log() log() log(( )!) < log + < log(!) e log(( )!) < e log + < e log(!) ( )! < e + <! e + <! <! e ( )! < e +! < + ( + )+ < e e 4

15 From these two separate iequalities we get what we wated: ( + )+ <! < e e 6. Prove that if a = l, the (a a ) = l. Hit: This problem is very similar to (i fact it is a special case of) Problem Proof: If ɛ > 0, we wat to fid a N > 0 such that (a+...+a) l < ɛ, N. So we have that a N+ l < ɛ, a N+ l < ɛ,..., a N+M l < ɛ ad if we put them all together we get a N+ l + a N+ l a N+M l < Mɛ Also, we ca say that a N+ + a N a N+M Ml < Mɛ As well we kow that ad a N+ + a N a N+M N + M a + a a N < ɛ N + M Ml N + M l < ɛ Ml N + M < Mɛ N + M < ɛ if we pick a appropriate N. Fially, if we put all three together we get that: a N+ + a N a N+M N + M Ml N + M + a + a a N N + M + Ml l < 3ɛ N + M a + a a N+M N + M l < 3ɛ Sice 3 is just a costat, we could go back ad adjust our ɛ s but it wo t chage the proof. 7. Suppose that f is cotiuous ad x f(x + ) f(x) = 0. Prove that x f(x)/x = 0. Hit: See the previous problem. Proof: Cosider the sequece of a = f( + ) f(). So we have x f(x + ) f(x) = x a = 0. From the problem above we get that (a a ) = 0 5

16 f( + ) f() + f() f( ) f() f() = 0 f( + ) f() = 0 + ) (f( f() ) = 0 f( + ) = 0 f() = 0 f() So for all itegers we kow that = 0 but that does t ecessarily imply that x f(x)/x eve if f(x) is cotiuous. So I am ot sure what to do from here. 8. Suppose that a > 0 for each ad that a + /a = l. Prove that a = l. Hit: This requires the same sort of argumet that works i Problem 6, together with the fact that a = l, for a > 0. a Proof: We kow that if a + /a = l the if + a a = sup + a = l. We a also kow that if + a if a sup a a sup + a, so by the Squeeze Theorem we have the desired result.. (a) Use Problem -5 to show that if c, the c m + c m c = cm c +. c Proof: It is easy to see that ( c)( + c + c c m ) = c m. So the we have: c m + c m c = c m ( + c + c c m ) = c m ( c m+ c We eed the fact that c because otherwise the right had side is udefied. (b) Suppose that c <. Prove that m, cm c = 0. ) = cm c +. c Proof: We kow that c < < c < c m < ad sice m <, we kow that c + < c m 0 < c m c + < ad we kow that 0 < c <. So the we have that 0 < cm c + c < (c) Suppose that {x } is a sequece with x x + c, where c <. Prove that {x } is a Cauchy sequece. 6

17 Proof: Lets try to use what we have from part (b) ad work toward the defiitio of a Cauchy Sequece. Assume < m, the we kow that x x m = x x + + x + x + + x x +m + x +m + x m x x m x x + + x + x x +m x +m + x +m + x m x x m c m c x x m m, m, cm c x x m 0 m, x x m < ɛ So we have that for all ɛ > 0 N such that, m > N x x m < ɛ so the sequece {x } is Cauchy. 3. Suppose that f is a fuctio o R such that ( ) f(x) f(y) c x y, forallxady, where c <. (Such a fuctio is called a cotractio). (a) Prove that f is cotiuous. Proof: We wat to show that ɛ > 0, δ > 0 such that x, y R, x y < δ f(x) f(y) < ɛ. If we pick δ = ɛ ɛ c ad from the coditio above o f, we get f(x) f(y) c x y < cδ = c( c ) = < ɛ. So f is cotiuous. ɛ (b) Prove that f has at most oe fixed poit. Proof: Assume there were two distict fixed poits x, y. The by defiitio f(x) = x ad f(y) = y f(x) f(y) = x y, but we also have that f(x) f(y) c x y where c < but this would imply that f(x) f(y) < x y which is a cotradictio. So f has at most oe fixed poit. (c) By cosiderig the sequece x, f(x), f(f(x)),... for ay x, prove that f does have a fixed poit. (This result, i a more geeral settig, is kow as the cotractio lemma. ) Proof: We kow that f(f(x)) f(x) c f(x) x because of the give property of f. As well, f(f(f(x))) f(f(x)) c f(f(x)) f(x) c f(x) x. So i geeral we have that f + (x) f (x) c f(x) x. So if we take the it as we get that f + (x) f (x) 0 f(f (x)) f (x). So the fixed poit is the it of the sequece above. 4. (a) Prove that if f is differetiable ad f <, the f has at most oe fixed poit. 7

18 Proof: Assume there were two distict fixed poits x, y. The by defiitio f(x) = x ad f(y) = y f(x) f(y) = x y f(x) f(y) x y = x y f(x) f(y) x y = x y f (y) =. This is a cotradictio to the fact that f < for all y. So that meas f has at most oe fixed poit. (b) Prove that if f (x) c < for all x, the f has a fixed poit. Proof: So f (x) = f(x) f(y) x y c < f(x) f(y) c x y. This meas that this is a cotractio. So if we cosider the sequece from questio 3 part (c), we will get the same coclusio that f has at least oe fixed poit, amely the it poit of the sequece. (c) Give a example to show that the hypothesis f (x) is ot sufficiet to isure that f has a fixed poit. Proof: Let f(x) = x +, the f (x) =. But this is a lie parallel to the lie f(x) = x so the two will ever cross, so f has o fixed poits. Actually ay lie of the form f(x) = x + c where c 0 will work. 5. This problem is a sort of coverse to the previous problem. Let b be a sequece defied by b = a, b + = f(b ). Prove that if b = b exists ad f is cotiuous at b, the f (b). Hit: If f (b) >, the f (b) > for all x i a iterval aroud b, ad b will be i this iterval for large eough. Now cosider f o the iterval [b, b ]. Proof: Followig the hit, assume f (b) >. Sice f is cotiuous, for all ɛ > 0 ad x (b ɛ, b + ɛ) f (x) >. The b = b meas there exists N such that b (b ɛ, b + ɛ) for all > N. That meas f is differetiable at b, which meas f is cotiuous at b. So from a problem above we have thatf(b) = b. Suppose b < b ad f > i this iterval, we have f(b) > f(b ). The f(b) f(b ) > f(b) f(b ) > b b b b + < b b b + < b b b This is a cotradictio to the assumptio that b < b + < b. Ad thigs would follow similarly if b > b ad f <. Also, for b = b, the f (b) = 0. Thus f (b). 7. Let {x } be a sequece which is bouded, ad let y = sup{x, x +, x +,...}. (a) Prove that the sequece of {y } coverges. The it y is deoted x or sup x, ad called the it superior, or upper it, of the sequece {x }. Proof: Sice {x } is bouded the {y } is bouded as well. Sice we are oly removig terms as the y is o-icreasig. So it coverges by the Mootoe Covergece Priciple. (b) Fid x for each of the followig: (i) x =. x = 0 8

19 (ii) x = ( ). x = 0 (iii) x = ( ) [ + ]. x = (iv) x =. x = (c) Defie x ad prove that Proof: Let {x } be as before, ad let x x. y = if{x, x +, x +,...}. The it y is deoted x or if x, ad called the it iferior, or lower it, of the sequece {x }. Also, we kow that y y. If we take the it as goes to ifiity of both sides we get: y y if{x, x +, x +,...} sup{x, x +, x +,...} x x. (d) Prove that x exists iff x = x ad that i this case x = x = x. Proof: ( ) Assume x exists. From above we have that x x. So it is eough to prove that x x. ( ) Assume x = x. This meas the smallest subsequetial it poit is equal to the largest subsequetial it poit. This meas there is oly oe. So that meas x exists ad is i fact equal to x = x (e) Recall the defiitio, i Problem 8-8, of A for a bouded set A. Prove that if the umbers x are distict, the x = A, where A = {x : N} 9