Solutions manual for introduction to analysis 5th by gaughan download

Chapter Sequences 1.1 Sequences and Convergence Show that [0; 1] is a neighborhood of { that is, there is > such that 32 ;3 + 1[0; 1] 12 1 Choose = Then 3 ; + = ; [0; 1] *2 Let x and y be distinct real numbers Prove there is a neighborhood P of x and a neighborhood Q of y such that P \ Q = ;: Choose = P \ Q = ;: jx yj Then, let P = (x ; x + ) and Q = (y ; y + ) Then, *3 Suppose x is a real number and > Prove that (x ; x + ) is a neighborhood of each of its members; in other words, if y (x ; x + ), then there is > such that (y ; y + ) (x ; x + ) jy (x+ )j Let = ; jy (x )j 2 Then, (y ; y + ) (x ; x + ) 3n+7 n n=1 Find upper and lower bounds for the sequence 3n+7 First, n = + n Thus, the lower bound is and the upper bound is 10 Give an example of a sequence that is bounded but not convergent n Let an = ( 1) Then, this sequence alternates between and 1, but never converges 13 14 Use the de nition of convergence to prove sequences converges: that each of the ( a) + 1n following (b ) n=1 n n=1 2n f gn= (c) 2n 3n n (b) Let > be gi ven Let N (d) o 2n+1 n=1 (a) Let > be given Let N = = n Then, n + = Then, n < + n1 = < n1 < N1 < N 2 2n+2n 2n Let (c) (d) lg j > be given Let N = = Then, n 2N < 2n +1 < > be given Let N Let j = = Then, = 3n = 4n+ < 4n < 4N < *7 Show that fang n=1 converges to A i fan Ag n=1 converges to CHAPTER SEQUENCES We see that j(an A) 0j = jan Aj < 1 Suppose fang n=1 converges to A, and de ne a new sequence fbng n=1 by bn = an+an+1 for all n Prove that fbng n=1 converges to A a ALet 2A jn j > be given We see that = an A+2a n+1A +j n+1 < + = Thus, j9 a s.t A N an+a n+1 bn A = 2n a n+ a +1 A ! 1 1 *9 Suppose fang n=1, fbn g n=1, and fcng n=1 such that fang n=1 converges to A, fbn g n=1 converges to A, and an cn bn for all n Prove that fcng n=1 converges to A Since an cn bn, we must have an A cn A bn A By convergence and de nition of absolute value, < an A cn A bn A < Hence, jcn Aj < Thus, cn ! A (We will call this result the Squeeze Theorem.) 1 *10 Prove that, if fang n=1 converges to A, then fjanjg n=1 converges to jAj Is the converse true? Justify your conclusion TI We see that jjanj jAjj < jjan Ajj = jan Aj < The converse is not true For instance, n n j( 1) j ! 1, but f( 1) g diverges *11 Let fang n=1 be a sequence such that there exist numbers and N such that, for n N, an = Prove that fang n=1 converges to We see that jan j jaN j = < for all > CHAPTER SEQUENCES 15 12 Give an alternate proof of Theorem 1.1 along the following lines Choose > There is N1 such that for n N1, jan Aj < , and there is N2 such that for n N2, jan Bj < Use the triangle inequality to show that this implies that jA Bj < Let N = max(N1; N2) Then, > jan Aj + jan Bj = jan Aj + jB anj > jan A + B anj = jB Aj Thus, jA Bj < 13 Let x be any positive real number, and de ne a sequence fangn =1 by an = n2 [x] + [2x] + + [nx] where [x] is the largest integer less than or equal to x Prove that fa ng n=1 converges to x=2 x(1+n + Let > be given and set N= x n) x = + xn2 2n n < 2x = xn(n2 2n xn2 2n ax +1) x2 Then, 2 x+2 xx+ n = =x +< nx n 2N 1.2 Cauchy Sequences 14 Prove that every Cauchy sequence is bounded (Theorem 1.4) Suppose that fang is not bounded Then, for any k, there is an nk such that jank j > k Then, fank gis an unbounded sequence Then, for any N, there exist x 16 ank and an` such that jank an` j > jank j jan` j = k ` where k ` > N Thus, fang is not Cauchy 1 15 Prove directly (do not use Theorem 1.8) that, if fang n=1 and fbng n=1 are Cauchy, so is fan + bng n=1 Since fang and fbng are Cauchy, then for all > 0, there exist N1 and N2 such that jan amj < for all m; n > N1 and jbn bmj < for all m; n > N2 Choose N = max(N1; N2) Then, jan + bn (am + bm)j = jan am + bn bmj < jan amj + jbn bmj < + = for all m; n > N 1 16 Prove directly (do not use Theorem 1.9) that, if fa n g n=1 and fbng n=1 are Cauchy, so is fanbng n=1 You will want to use Theorem 1.4 Since fbng is Cauchy, then it is bounded (by Exercise 14) Thus, jb nj < M for some M Since fang is Cauchy, then for all > 0, there exist N ja n am j < M for all m; n > N and jbn bmj < for all m; n > N2 Let > be given Then, janbn ambmj < janM amMj = jM(an am)j < CHAPTER 2n+1 sequence nm n m n 17 n = SEQUENCES n=1 < N2 mn mn n that the 2mn+m 2mn = N < Then, = Let > be given Choose N = m Prove is Cauchy N 2n+1 n m 2m+1 = nm 18 Give an example of a sequence with exactly two accumulation points n if n is even 1=+ 1=n ;; if n is odd Then, an has accumulation points at Let an = and 19 Give an example of a set with a countably in nite set of accumulation points CHAPTER SEQUENCES 17 The set Q has the property that every element is an accumulation point, a a a since for any b Q, the sequence b + n converges to b Since Q is countable, we have found the desired set 20 Give an example of a set that contains each of its accumulation points The set [0; 1] contains all of its accumulation points 21 Determine the accumulation points of the set 2n + k1 : n and k are positive integers + n n The set f2 : n Z g[f1g is the set of accumulation points since + k ! n n as k ! and + k ! as n ! 22 Let S be a nonempty set of real numbers that is bounded from above (below) and let x = sup S (inf S) Prove that either x belongs to S or x is an accumulation point of S It is clear that x S is a possibility Suppose x 2= S Then, by Exercise 0.44, for any > 0; there is an a S such that x < a < x Thus, for all n, there 1 exists an an S such that x n < an < x Since x n ! x, we have an ! x Thus, x is an accumulation point of S 23 Let a0 and a1 be distinct real numbers De ne an = an 1+an for each2 positive integer n Show that fang n=1 is a Cauchy sequence You may want to use induction to show that n an+1 an = (a1 a0) and then use the result from Example 0.9 of Chapter The statement an+1 an = Suppose that it is true for n < n (a1 a0) is obviously true for n = N Then, a a = a +a N+1 h N a N +1 +aN aN ( N N a a )+ (a a ) = (a = i 1 a)+a + 18 N+1 1 N N N + (a1 a)= +1 (a N+1 i h +1 n N+2 N a)= N Thus, the s tatement is proven by i nducti on N+1 i N a ) + (a i N + a0) = h (a a = ) l m N +1 (a a)= Now, let > and n; m be given Choose N = lg ja1 a0j(n m) Then, a +a2 j + (a1 a0 ) an + + = +< + a an + + am+1 jan am j = am < n m n m a n m aN m +1for +1 aN 1(a j j n jj j n N: Thus, we see that am < an n n n m 24 Suppose fang n=1 converges to A and fan : n Ng is an in nite set Show that A is an accumulation point of fan : n Ng k for all n fn Ng k A k ; a + k By convergence, there is an N such that jan Aj < there is an element of a : n in N for every k Now, every neighborhood of A has Nk as a subset for some k and since there are an Let Nk = N Thus, in nity of Nk's, we have an in nity of members of fang in any neighborhood 1.3 Arithmetic Operations on Sequences 1 25 Suppose fang n=1 and fbng n=1 are sequences such that fang n=1 and 1 fan + bng n=1 converge Prove that fbng n=1 converges Suppose an ! A and an + bn ! C Then, fbng = fan + bn ang Thus, by Theorem 1.8, bn ! C A 1 26 Give an example in which fang n=1 and fbng n=1 not converge but fan + bng n=1 converges CHAPTER SEQUENCES n Let an = ( 1) and bn = ( 1) 19 n+1 We know that an and bn don't converge, but an + bn ! 1 27 Suppose fang n=1 and fbng n=1 are sequences such that fang n=1 con1 verges to A 6= and fanbng n=1 Prove that fbng n=1 converges n C anbn o Suppose anbn ! C Then, fbng = 28 If a tofpnag.n1=1 , so by Theorem 1.9, bn ! A an for all n, show converges to a with a pan n n=1 converges CHAPTER SEQUENCES 18 j Let > be given Then, there is an N such that p n j a jn p ana +n pa a= a jn pan +p a =a a j j a a

> > < > (n + k) > = k : k > > n=1 ; converges to , where k ! n + k = (n + k)! : k n!k! We see that 1)( n+k k2) (n+1) k!(n +k) n!k1 o k n o (n+!(kn)!+ k) =n k (n+ k Now, let > be 1) ( n+k k (n+1) (n + k)k (n+1) k 1 given Choose N = Then, ) n k k! = < k!(n+k) k! = n+1 = n+1 N k k!(n+k1 k < k!(n+ k) k! k!(n+ k) k!(n+ k) < for all n n 30 Prove the following variation on Lemma 1.10 If fbng n=1 converges to B 6= and bn 6= for all n, then there is M > such that jbnj M for all n Choose M = jbnj =2 Then, the statement holds 31 Consider a sequence fang n=1 and, for each n, de ne a+a++a n= :n n 1 Prove that if fang n=1 converges to A, then f ng n=1 converges to A Give an 1 example in which f ng n=1 converges, but fang n=1 does not Let > be given There is an N1 such that jan Aj Let M = ja1 Aj + ja2 Aj + + jaN1 < for all n > N Aj Then, there is an N2 such that < for all n > N Let N = max(N ; N ) Then, a1+a2+ + an A=n M n a 2j n nA + +aN + +an 2+2= j of convergence, a n n < n A + +jaN Aj + jaN+1 Aj+ +jan Aj = M + jaN+1 Aj+ +jan Aj to show this would be to observe that by the de nition < for all n > N for some N Then, since nj 1)n Then, as we have seen before, a g Let a = ( 0: < a1 (A quicker way j n diverges, but a1+a2+ f n +a n n ! n 32 Find the limit of the sequences with general term as given: (a) n2+4 n (b) cos n n5 n sin n (c) pn CHAPTER SEQUENCES 19 n (d) n2 ( e) q n pn n1 n (f) ( 1) n+7 (a) p (b) (c) (d) 0 2n = n2 q (e) q 1 n n n = n n 4n = Thus, limit is (f) n2 n p 4n n+ 4n2 p 4n n p n2 2 33 Find the limit of the sequence in Exercise 23 when a0 = and a1 = You might want to look at Example 0.10 must be 2 2n ! n Example 0.10 says that limit 1.4 a +a n n = + Since Subsequences and Monotone Sequences 34 Find a convergent subsequence of the sequence 1 0, the CHAPTER SEQUENCES 20 ( 1)n n Let nk = 2n Then, the subsequence is n=1 n which converges to 35 Suppose x is an accumulation point of fan : n Ng Show that there is a subsequence of fang n=1 that converges to x Since x is an accumulation point, every neighborhood about x contains an in nity of fang Thus, let ank be a member of fan : n Ng \ x k ; x 1 + k Then, for any > 0, there is a K such that k < for all k > K Thus, ank ! x 36 Let fang n=1 be a bounded sequence of real numbers Prove that fang n=1 has a convergent subsequence Eitherfang has a nite number of values or fang has an in nite number of values For the former, there must be some value x for which there are in nitely many k such that ank = x Thus, ank ! x For the latter, the sequence is a bounded in nite set of real numbers, so by the Bolzano-Weierstrass Theorem, an has a convergent subsequence 1 *37 Prove that if fang n=1 is decreasing and bounded, then fang n=1 converges Assume that fang attains an in nite number of values Suppose that inf an = > be given Then, there are a1 M intervals that sequence values M Let may fall Since this is a nite number and there are an in nite number of values, at least one region must contain an in nite number of function values Since the sequence is decreasing, the last region must contain an in nity of values; that is, an (M; M + ) for all n > N for some N Since was arbitrarily chosen, the proof is complete The case for when fang has only nitely many values is easy n 38 Prove that if c > 1, then f It is clear that f Also, pc n g pcgn1=1 converges to n pnc > is a monotone decreasing sequence c > 1, so by Theorem 1.16, c ! n+1 n pp pc Thus, CHAPTER SEQUENCES 21 1 *39 Suppose fxng n=1 converges to x0 and fyng n=1 converge to x0 De ne a sequence fzng n=1 as follows: z2n = xn and z2n = yn Prove that fzng n=1 converges to x0 Both subsequences of fzng converge to x0 Thus, by Theorem 1.14, zn ! x0 p 40 Show that the sequence de ned by a1 = and an = + an for n > is convergent and nd its limit p 2 To nd the limit L, set L = + L , L = + L , L L = The solutions are and The only solution that works is Thus, a ! We p p p n provepthat fan g is decreasing Since + an < + an and we know that an is decreasing, we see that the whole sequence is decreasing Also, square roots must be greater than 0, so the sequence is bounded Thus, the sequence is bounded below and decreasing and is thus convergent 41 Let fxng n=1 be a bounded sequence and let E be the set of subsequential limits of that sequence By Exercise 36, E is nonempty Prove that E is bounded and contains both sup E and inf E Since fxng is bounded (by M), its limit points must be such that they are within distance of some sequence values Thus, limit points must be within the same bounds as fxng or within distance of the boundary for any Thus, E is bounded (by, say M + 1) We must ensure that members of E not form a sequence themselves that converges to a non-limit point So, suppose there is a sequence feng of limit points Then, for every , there is an N such that jen xnj < for all n > N Thus, xn ! en Thus, all sequences of E converge in E ( since they are estimated by subsequences of fx ng Thus, sup E; inf E E 42 Let fxng n=1 be any sequence and T : N ! N be any 1-1 function Prove that if fxng1n=1 converges to x, then fxT (n)g1n=1 also converges to x Explain how this relates to subsequences De ne what one might call a \rearrangement" of a sequence What does the result imply about rearrangements of sequences? We see that fxT (n)g = fxn1 ; xn2 ; :::g and is a subsequence of fxng Since all subsequences converge, we must have xT (n) ! x Let T : N ! N be any 1-1 function and let fxng be a sequence Then, fxT (n)g is called a rearrangement The result implies that if fxng converges, then so does fxT (n)g for any T CHAPTER SEQUENCES 22 43 Assume a b Does the sequence f(a diverge? If the sequence converges, nd the limit n n 1=n + b ) g n=1 converge or The sequence does not converge in general For instance, if a = and b = n 1=n n 1, then the sequence becomes f[1 + ( 1) ] g Taking even indexes, the limit is 1, and taking odd indexes, the limit is Thus, not all subsequences converge to the same limit point, so the sequence is not convergent 44 Does the sequence P ) (k pk n=1 +n k=1 X diverge or converge? If the sequence converges, nd the limit k 1 1 k We see that > n=1 p k1 +k =pP n=1 p k k 2+ k k2+n < n=1 k =k = Also, n=1 p k2+n ! Thus, the sequence must converge to by theP P Squeeze Theorem (established in Exercise 9) *45 Show that if x is any real number, there is a sequence of rational numbers converging to x Let a0:a1a2 be the decimal expansion for x Then, de ne xn = a0:a1a2 an Then, xn Q for all n and xn ! x (for there exists an N for which xn can be within k 10 distance for any integer k and n > N) *46 Show that if x is any real number, there is a sequence of irrational numbers converging to x If x is already irrational, de ne x n x Clearly, xn ! x If x is rational, de ne x n = x + n Then, xn R n Q for all n and xn ! x CHAPTER SEQUENCES 22 47 Suppose that fang n=1 converges to A and that B is an accumulation point of fan : n Ng Prove that A = B 1 If B is an accumulation point, then B n ; B + n contains a member of fang for all n Thus, one can construct a subsequence of these members that converges to B, and since fang is convergent, we must have A = B Miscellaneous 1 48 Suppose that fang n=1 and fbng n=1 are two sequences of positive real numbers We say that an is O(bn) (read as \big oh" of bn) if there is an integer N and a real number M such that for n N, an M bn Prove that if fan=bng n=1 converges to L 6= 0, then an is O(bn) and bn is O(an) What can you say if L = 0? Illustrate with examples Since an=bn ! L, we guess that there is an N such that an (L + 1) bn for all n N We now prove this assertion First, for any > 0, there is an N for which an=bn (L ; L + ) for all n N Thus, for the same N, an (bn[L ]; bn[L + ]) Thus, an bn(L + 1) Thus, an is O(bn) if N is chosen to correspond to = Similarly, bn is O(an) If L = 0, then either n ! n a bounded For instance, n3 ! and LimitComparison Test and g f b n is bounded or b n ! 1= n 2+1=n and a fn g is = (This result is called the