Introduction to analysis maxwell rosenlicht

Chapter Sequences Download Full Solutions manual Introduction to Analysis Maxwell Rosenlicht https://getbooksolutions.com/download/solutions-manual-introduction-to-analysis 1.1 Sequences and Convergence Show that [0; 1] is a neighborhood of 2 ;3 + 1[0; 1] 2 2{ that is, there is > such that 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 = jx yj Then, let P = (x P \ Q = ;: ; 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 = 3n+7 jy (x )j ; Then, (y ; y + ) (x ; x + ) 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 5 Give an example of a sequence that is bounded but not convergent Let an = ( 1)n Then, this sequence alternates between and 1, but never converges 13 CHAPTER SEQUENCES 14 Use the de nition of convergence to prove that each of the following sequences converges: (a) + 1 (b) f (c) (d) n n n n=1 2n n=1 gn=1 n 3n 2n+1 o n=1 (b) Let > be given Let N (a) Let > be given Let N = 2 n < (c) (d) Let N Let = Then, Then, < n + n1 > be given Let N = = Then, Then, j 2n+1 3n = = n j n < 2n N = < 2n+2n n lg > be given Let N + 2n = < = 2N < 6n 6n 4n+2 = < 4n+2 4n < 4N < 3 *7 Show that fang1n=1 converges to A i fan We see that j(an A) 0j = jan Ag1n=1 converges to Aj < Suppose fang1n=1 converges to A, and de ne a new sequence fbng1n=1 by bn = an+an+1 for all n Prove that fbng1n=1 converges to A a A a A N Let > be given We see that < j = n n+1 n+1 + = Thus, = a +a 2A n n+1 an A+an+1A bn A ! *9 Suppose fang1n=1, fbn g1n=1, and fcng1n=1 such that fang1n=1 converges to A, fbn g1n=1 converges to A, and an cn bn for all n Prove that fcng1n=1 converges to A jn j +j A a +a s.t 2 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.) *10 Prove that, if fang1n=1 converges to A, then fjanjg1n=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, j( 1)nj ! 1, but f( 1)ng diverges *11 Let fang1n=1 be a sequence such that there exist numbers and N such that, for n N, an = Prove that fang1n=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 fangn1=1 by an = [x] + [2x] + + [nx] n2 where [x] is the largest integer less than or equal to x Prove that fang1n=1 converges to x=2 Let > be given and set N = x(1+ +n) n x = xn(n2 x 2n = 2n x xn2 22 Then, a + xn2 2n x x n = 2 x 2n < x+2x+2 +nx x < 2N n x = +1) 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 ank and an` such that jank an` j > jank j jan` j = k ` where k ` > N Thus, fang is not Cauchy 15 Prove directly (do not use Theorem 1.8) that, if fang1n=1 and fbng1n=1 are Cauchy, so is fan + bng1n=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 16 Prove directly (do not use Theorem 1.9) that, if fan g1n=1 and fbng1n=1 are Cauchy, so is fanbng1n=1 You will want to use Theorem 1.4 Since fbng is Cauchy, then it is bounded (by Exercise 14) Thus, jbnj < 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 SEQUENCES 16 2n+1 17 Prove that the sequence nm = < mn mn N2 = N Let > be given Choose N = m n m n N n < is Cauchy n=1 Then, 2n+1 n 2m+1 m = 2mn+m 2mn n nm = 18 Give an example of a sequence with exactly two accumulation points Let an = and n ; = if n is even + 1=n ; if n is odd Then, an has accumulation points at 19 Give an example of a set with a countably in nite set of accumulation points The set Q has the property that every element is an accumulation point, since for any ab Q, the sequence ab + n converges to ab 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 + The set f2n : n Z g[f1g is the set of accumulation points since 2n + k ! 2n as k ! and 2n + 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 exists an 1 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 each positive integer n Show that fang1n=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 CHAPTER SEQUENCES The statement an+1 17 an = (a1 a0) is obviously true for n = N Then, a a = a +a N+1 Suppose that it is true for n < h n N (a a ) = (a1 N+1 N 1 N N = a0) + aN N a )= aN+1+aN 2aN (a a )+ N+1 (a1 a h N N + N N 2 i +1 (a a )= N Thus, the s tatement is proven by i nducti on (a1 a0) Then, a +a a n n n a + + + n m = m l be given Choose N = n lg ja1 a0j(n m) j j jj j jan amj = < n m for n N: Thus, we see that a + +a am < an n m+1 am < + + am+1 n m 24 Suppose fang1n=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 N Thus, Let Nk = f n Ng k A k ; a + k By convergence, there is an N such that ja n 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 in nity of Nk's, we have an in nity of members of fang in any neighborhood 1.3 (a i N+1 a )= + a0) = h (a Now, let > and n; m n m a a)= i n j + + (a 2 a ) + (a N+1 N+2 N+1 i Arithmetic Operations on Sequences 25 Suppose fang1n=1 and fbng1n=1 are sequences such that fang1n=1 and fan + bng1n=1 converge Prove that fbng1n=1 converges Suppose an ! A and an + bn ! C Then, fbng = fan + bn ang Thus, by Theorem 1.8, bn ! C A 26 Give an example in which fang1n=1 and fbng1n=1 not converge but fan + bng1n=1 converges Let an = ( 1)n and bn = ( 1)n+1 We know that an and bn don't converge, but an + bn ! 27 Suppose fang1n=1 and fbng1n=1 are sequences such that fang1n=1 con-verges to A 6= and fanbng1n=1 Prove that fbng1n=1 converges n o C anbn Suppose anbn ! C Then, fbng = an 28 If a converges to a with an f g converges to pna n1=1 , so by Theorem 1.9, bn ! for all n, show pan A n=1 CHAPTER SEQUENCES 18 j Let > be pgiven Then, there is an N such that p p n a an+ a= an a a j jn =a 29 Prove that n+k > > > : = (n + k)k > k > ; n=1 converges to k ! , where j a > > < j n pan+pa a j an a

be (n+k)k 2(n+1) = k k!(n+k) k! < k!(n +k) o < ) k k!(n+k k = n+1 n k k! (n+1) k! N < fo r all n n 30 Prove the following variation on Lemma 1.10 If fbng1n=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 fang1n=1 and, for each n, de ne n= a +a + +a n :n Prove that if fang1n=1 converges to A, then f ng1n=1 converges to A Give an example in which f ng1n=1 converges, but fang1n=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 Let N = max(N ; N ) Then, M n a 2j n nA + +aN + +an < a1 j n < for all n > N Aj Then, there is an N2 such that A =n a1+a2+ +an n n < n A + +jaN Aj + jaN+1 Aj+ +jan Aj = M + jaN+1 Aj+ +jan Aj + = (A quicker way to show this would be to observe that by the de nition j j n of convergence, a < for all n > N for some N Then, since a +a + +a Let a = ( 1)n Then, as we have seen before, a g diverges, but n f n n 0: 32 Find the limit of the sequences with general term as given: (a) n +4n cos (b) n n2 n5 sin n2 (c) pn n ! CHAPTER SEQUENCES n (d) n2 q n pn n (f) ( 1) n+7 (e) (a) (b) (c) (d) 19 n 0 q q 1 2n = n (e) n n = n n 4n Thus, limit is (f) p 4n2 n p 4n2 = 4n 4n p n 4n n+ 4n p 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 2n 2n ! a +a n n Example 0.10 says that = + Since 0, the limit 1.4 Subsequences and Monotone Sequences 34 Find a convergent subsequence of the sequence ( 1)n n Let nk = 2n Then, the subsequence is 1 n=1 n which converges to 35 Suppose x is an accumulation point of fan : n Ng Show that there is a subsequence of fang1n=1 that converges to x Since x is an accumulation point, every neighborhood about x contains an 1 in nity of fang Thus, let ank be a member of fan : n Ng \ x k ; x + k Then, for any > 0, there is a K such that k < for all k > K Thus, ank ! x 36 Let fang1n=1 be a bounded sequence of real numbers Prove that fang1n=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 CHAPTER SEQUENCES 20 bounded in nite set of real numbers, so by the Bolzano-Weierstrass Theorem, an has a convergent subsequence *37 Prove that if fang1n=1 is decreasing and bounded, then fang1n=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 38 Prove that if c > 1, then f It is clear that n p n pc > n+1 p n cgn1=1 converges to n pc Thus, pc n p f g is a monotone decreasing sequence Also, c > 1, so by Theorem 1.16, c ! *39 Suppose fxng1n=1 converges to x0 and fyng1n=1 converge to x0 De ne a sequence fzng1n=1 as follows: z2n = xn and z2n = yn Prove that fzng1n=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 = 6+L,L =6+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 fxng1n=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 CHAPTER SEQUENCES 21 jen xnj < for all n > N Thus, xn ! en Thus, all sequences of E converge in E (since they are estimated by subsequences of fxng Thus, sup E; inf E E 42 Let fxng1n=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 fx ng Since all subsequences converge, we must have x T (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 1=n 43 Assume a b Does the sequence f(an + bn) g n=1 converge or diverge? If the sequence converges, nd the limit The sequence does not converge in general For instance, if a = and b = 1=n 1, then the sequence becomes f[1n + ( 1)n] 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 p ) k2 + n n=1 k=1 X diverge or converge? If the sequence converges, nd the limit We see that =p p n=1 k n=1 k p 1 ! k +k P P Squeeze Theorem (established in Exercise 9) 2+k k k 1 < = k = Also, > n=1 k n=1 p k +n Thus, the sequence must converge to by the 2 k +n P *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 10k 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 xn x Clearly, xn ! x If x is rational, de ne xn = x + n Then, xn R n Q for all n and xn ! x CHAPTER SEQUENCES 22 47 Suppose that fang1n=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 48 Suppose that fang1n=1 and fbng1n=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=bng1n=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) a and b g is bounded or b and a is f n n!1 If L = 0, then either n ! f ng bounded For instance, n3 ! and 2+1=n = (This result is called the Limit Comparison Test n 1=n