= 1 lim sup{ sn : n > N} )

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1 ATH 104, SUER 2006, HOEWORK 4 SOLUTION BENJAIN JOHNSON Due July 12 Assgmet: Secto 11: 11.4(b)(c), 11.8 Secto 12: 12.6(c), 12.12, Secto 13: 13.1 Secto Cosder the sequeces s = cos ( ) π 3, t = 3, u 4+1 = (, 2) 1 v = ( 1) + 1. (b) For each sequece, gve ts set of subsequetal lmts. w = ( 2) : {, } x = 5 ( 1) : { 1, 5} 5 y = 1 + ( 1) : {0, 2} z = cos ( ) π 4 : {, 0, } (c) For each sequece, gve ts lm sup ad lm f. w = ( 2) : lm f =, lm sup = x = 5 ( 1) : lm f = 1, lm sup = 5 5 y = 1 + ( 1) : lm f = 0, lm sup = 2 z = cos ( ) π 4 : lm f =, lm sup = 11.8 (a) Use Defto 10.6 ad Exercse 5.4 to prove that lm f s = lm sup( s ) Proof. For each N N we have f{s : > N} = sup{ s : > N} (by exercse 5.4). Hece lm (f{s : > N}) = lm ( sup{ s : > N} ). Usg Defto 10.6, we have lm f s = lm (f{s : > N}) ( = lm sup{ s : > N} ) ( = 1 lm sup{ s : > N} ) = lm sup s (b) Let t k be a mootoc subsequece of s covergg to lm f s. Show that t k s a mootoc subsequece of s covergg to lm f s Observe that ths completes the proof of Corollary Proof. Sce t k s a subsequece of s, for each k there s k such that 1 < 2 < < k < k+1 < ad t k = s k. Sce t k = s k s equvalet to t k = s k, we have t k s a subsequece of s as well. Date: July 12,

2 2 BENJAIN JOHNSON Secto 12 Sce t k s mootoe, ether ( k N)(t k t k+1 ) or ( k N)(t k t k+1 ). I the frst case, ( k N)( t k t k+1 ), ad the secod case, ( k N)( t k t k+1 ). So t k s mootoe as well. Fally, sce t k lm f s, we have t k 1 lm f s = lm sup( s ). The proof of Corollary 11.4 showed that for every sequece s, there s a mootoe subsequece of s covergg to lm sup s. Let s be ay sequece, ad apply the Corollary to s. Ths produces a subsequece t k of s wth t k lm sup( s ). Ths exercse the shows that t k s a subsequece of s wth lmt lm f s. So for every sequece s, there s also a mootoe subsequece of s covergg to lm f s Let s be a bouded sequece, ad let k be a o-egatve real umber. (c) What happes (a) ad (b) f k < 0? If k < 0, the (a) becomes lm sup(ks ) = k lm f s, ad (b) becomes lm f(ks ) = k lm sup s. Proof usg (a) ad (b) for postve k: ad lm sup(ks ) = lm f( ks ) = ( k) lm f s = k lm f s lm f(ks ) = lm sup( ks ) = ( k) lm sup s = k lm sup s Let s be a sequece of o-egatve real umbers, ad for each defe σ = 1 (a) Show that lm f s lm f σ lm sup σ lm sup s. Proof. Let > > N N. The we have ad we have σ = 1 s =1 s + 1 s =1 =N+1 < 1 N s + 1 N sup{s k : k > N} =1 =N+1 s + sup{s k : k > N} =1 ( =1 s ).

3 ATH 104, SUER 2006, HOEWORK 4 SOLUTION 3 N σ = 1 s =1 s + 1 s =1 =N+1 > f{s k : k > N} = N So for each > > N, we have N =N+1 f{s k : k > N} > N f{s k : k > N} f{s k : k > N} < σ < 1 Sce was arbtrary teger greater tha, we have N s + sup{s k : k > N} f{s k : k > N} f{σ : > } sup{σ : > } 1 =1 N s + sup{s k : k > N} Now the above expresso cotas oly the free varables ad N, ad t holds for every > N. Takg the lmt as yelds f{s k : k > N} lm f σ lm sup σ sup{s k : k > N} Sce the above holds for every N, takg the lmt as N yelds =1 lm f s lm f σ lm sup σ lm sup s (b) Show that f lm s exsts, the lm σ exsts ad lm s = lm σ. Proof. If lm s exsts, the (by Theorem 10.7) lm f s = lm sup s = lm s. It s the clear from part (a) that all the equaltes must be equaltes. I partcular, lm f σ = lm sup σ. So (aga by Theorem 10.7) lm σ exsts, ad shares the commo value of all the relevat lm ad lm f expressos. I partcular lm s = lm σ Let s be a bouded sequece R. Let A be the set of a R such that { N : s < a} s fte. Let B be the set of b R such that { N : s > b} s fte. Prove that sup A = lm f s ad f B = lm sup s. (1) sup A lm f s Proof. Let a A. The { N : s < a} s fte. Choose = max{ : s < a}. (Ca do ths sce there are oly ftely may such ). The for every > N we have s a. So for each N, f{s : > N} a. Takg the lmt as N yelds lm f s a. Sce a was arbtrary A, ths argumet shows that lm f s s a upper boud for A. Sce sup A s the least upper boud for A, sup A lm f s.

4 4 BENJAIN JOHNSON Secto 13 (2) lm f s sup A Proof. Let ɛ > 0. I clam that (lm f s ɛ) A. If ot, the the set S = { N : s < (lm f s ɛ)} s fte. Cosder ay subsequece cosstg oly of terms S. By B.W., ths sequece has a coverget subsequece, whch s also a subsequece of s. Sce all the terms of ths subsequece are at most (lm f s ɛ), ts lmt L must satsfy L (lm f s ɛ). But ths cotradcts the theorem that says lm f s s the f of the set of all subsequetal lmts. We coclude that (lm f s ɛ) A. Sce sup A s a upper boud for A, lm f s sup A. (3) f B lm sup s Proof. Let ɛ > 0. I clam that (lm sup s + ɛ) B. If ot, the the set S = { N : s >) lm sup s + ɛ)} s fte. Cosder ay subsequece cosstg oly of terms S. By B.W., ths sequece has a coverget subsequece, whch s also a subsequece of s. Sce all the terms of ths subsequece are at least (lm sup s + ɛ), ts lmt L must satsfy L (lm sup s + ɛ). But ths cotradcts the theorem that says lm sup s s the sup of the set of all subsequetal lmts. We coclude that (lm sup s + ɛ) B. Sce f B s a lower boud for B, f B lm sup s. (4) lm sup s f B Proof. Let b B. The { N : s > b} s fte. Choose = max{ : s > b}. (Ca do ths sce there are oly ftely may such ). The for every > N we have s b. So for each N, sup{s : > N} b. Takg the lmt as N yelds lm sup s b. Sce b was arbtrary B, ths argumet shows that lm sup s s a lower boud for B. Sce f B s the greatest lower boud for B, lm sup s f B For pots x, y R k, let d 1 (x, y) = max{ x y : = 1, 2,..., k} ad d 2 (x, y) = k =1 x y. (a) Show that d 1 ad d 2 are metrcs for R k. Proof. We must show d 1 ad d 2 satsfy D1, D2, ad D3. Let x R k. d 1 (x, x) = max{ x x : = 1, 2,..., k} = 0 ad d 2 (x, x) = k =1 x x = 0. So d 1 ad d 2 satsfy D1. Let x, y R k. The d 1 (x, y) = max{ x y : = 1, 2,..., k} = max{ y x : = 1, 2,..., k} = d 1 (y, x) ad d 2 (x, y) = k =1 x y = k =1 y x = d 2 (y, x). So d 1 ad d 2 satsfy D2. Let x, y, z R k. The d 1 (x, z) = max{ x z : = 1, 2,..., k} max{ x y : = 1, 2,..., k} + max{ y z : = 1, 2,..., k} = d 1 (x, y) + d 1 (y, z), ad d 2 (x, z) = k =1 x z k =1 x y + k =1 y z = d 2 (x, y) + d 2 (y, z). So d 1 ad d 2 satsfy D3. (b) Show that d 1 ad d 2 are complete.

5 ATH 104, SUER 2006, HOEWORK 4 SOLUTION 5 d 1 Proof. Suppose that x () s a sequece from R k that s Cauchy wth respect to d 1. We frst show that for each k, x () s a Cauchy sequece wth respect to the usual metrc o R. Let ɛ > 0. Sce x s a cauchy sequece wth respect to d 1, there s N 0 R such that ( m, N : m > > N 0 )(d 1 (x (), x (m) ) < ɛ). Note: d 1 (x (), x (m) ) < ɛ meas max{ x () : = 1, 2,..., k} < ɛ Choose N = N 0. Let N. Assume > N. Let k. The x () max{ x () x (m) : = 1, 2,..., k} < ɛ. Ths shows that for each, x () s a Cauchy sequece wth respect to the usual dstace metrc o R. Sce R s complete wth respect to the usual metrc, each x () coverges to some real umber L. Now we show that L = (L 1, L 2,..., L k ) s the lmt of x (). Let ɛ > 0. Sce each x () coverges to some y. There exst N 1,... N k such that for each, > N k x () L < ɛ. Choose N = max{n 1,..., N k }. Let N wth > N. The d 1 (x (), L) = max{ x () L : = 1, 2,..., k} < ɛ. d 2 Proof. Suppose that x () s a sequece from R k that s Cauchy wth respect to d 2. We frst show that for each k, x () s a Cauchy sequece wth respect to the usual metrc o R. Let ɛ > 0. Sce x s a cauchy sequece wth respect to d 1, there s N 0 R such that ( m, N : m > > N 0 )(d 2 (x (), x (m) ) < ɛ). Note: d 2 (x (), x (m) ) < ɛ meas k =1 x () < ɛ Choose N = N 0. Let N. Assume > N. Let k. The x () k =1 x () x (m) < ɛ. Ths shows that for each, x () s a Cauchy sequece wth respect to the usual dstace metrc o R. Sce R s complete wth respect to the usual metrc, each x () coverges to some real umber L. Now we show that L = (L 1, L 2,..., L k ) s the lmt of x (). Let ɛ > 0. Sce each x () coverges to some y. There exst N 1,... N k such that for each, > N k x () L < ɛ. Choose N = max{n k 1,..., N k }. Let N wth > N. The d 2 (x (), L) = k =1 x () L < k =1 ɛ = ɛ. k