Solutions manual for introduction to analysis 5th by gaughan download

Give an example of a sequence that is bounded but not convergent.. Prove that, if fang1n=1 converges to A, then fjanjg1n=1 converges to jAj.. Prove that every Cauchy sequence is bounded.

Chapter 1 Sequences

1.1 Sequences and Convergence

1 Show that [0; 1] is a neighborhood of { that is, there is > 0 such that

32 ; 3 + 1[0; 1] 2 1 2 1 1

Choose = 3 Then 3 3 ; 3 + 3= 3 ; 1 [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 = jx2yj Then, let P = (x ; x + ) and Q = (y ; y + ) Then,

P \ Q = ;:

*3 Suppose x is a real number and > 0 Prove that (x ; x + ) is a neighborhood of each of its members; in other words, if y 2 (x ; x + ), then there

is > 0 such that (y ; y + ) (x ; x + )

jy (x+ )j jy (x )j

; 2 Then, (y ; y + ) (x ; x + )

4 3n+7 7 n n=1

Find upper and lower bounds for the sequence 3n+7 1 . First, n = 3 + n Thus, the lower bound is 3 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 1 and 1, but never converges

14

6 Use the de nition of

convergence to prove

that each of the

following

Let lg1 Then, 2 n 1

(c) > 0 be given Let N = j = < 2N <

(d) Let > 0 be given Let N = 3 4Then, j 2n+1 3n =

=

4n+ 2 3 < 4n 3 <4N 3 <

*7 Show that fang1n=1 converges to A i fan Ag1n=1 converges to 0 sequences converges: ((b ) a) 5 + n1n 1n=1 n=1 2 2n 1

f

gn= 1 (c) 2 n 1

3n 1

n o (b) Let > 0 be gi ven Let N (d) 2n+1 n=1

(a) Let > 0 be given Let N =

=

n

= Then, n + =

2

< 1 Then, 5 + n15 = n1 < N1 2 n < 2 <

N

2 2

2n

2 2n+2n

We see that j(an A) 0j = jan Aj <

8 Suppose fang1n=1 converges to A, and de ne a new sequence fbng1n=1 by

b n = an+a n+1

for all n Prove that fb n g 1

n=1 converges to A.2

a A Let > 0 be given We see that a A N a n+a n+1 2 A = a n+ 2a n+1

2A = a n A+ 2an+1A

j n j + j n+1 j 9< + = Thus, bn A

2 2 s.t 2 2 !

*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

Since an cn bn, we must have an A cn A bn A By convergence and de nition

of absolute value, < an1 A cn A bn A < Hence, jcn Aj < Thus, cn ! A (We will

*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 = 0 < for all > 0

CHAPTER 1 SEQUENCES 15

12 Give an alternate proof of Theorem 1.1 along the following lines

Choose > 0 There is N1 such that for n N1, jan Aj < 2 , and there is N2 such

that for n N2, jan Bj < 2 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 = [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 > 0 be given and set

N =

x2 = 2n xn2x2 2 Then,

x

2

n < x+2x x+ 2 n +< nx

2x

=

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

jan j > k Then, fan gis an unbounded sequence Then, for any N, there exist

16

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 < 2 for all m; n > N1 and jbn bmj < 2 for all m; n > N2

Choose N = max(N1; N2) Then, jan + bn (am + bm)j = jan am + bn bmj < jan amj + jbn bmj < 2 + 2 = 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 jan am j < M

for all m; n > N and jbn bmj < 2 for all m; n > N2 Let > 0 be given Then, janbn

ambmj < janM amMj = jM(an am)j <

sequence n n=1 is Cauchy

nm = mn mn < N2 = N<

1 Then, 2n+1 2m+1 = 2mn+m 2mn

n = Let > 0 be given Choose N =

18 Give an example of a sequence with exactly two accumulation points

1

n if n is even

Let an = 1 = + 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

CHAPTER 1 SEQUENCES 17

The set Q has the property that every element is an accumulation point, since for any a 2 Q, the sequence a + n converges to a 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 2 Z+g[f1g is the set of accumulation points since 2n + k1 ! 2n

as k ! 1 and 2n + k1 ! 1 as n ! 1

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 2 S is a possibility Suppose x 2= S Then, by Exercise

0.44, for any > 0; there is an a 2 S such that x < a < x Thus, for all n, there

exists an an 2 S such that x n < an < x Since x n ! x, we have an ! x Thus, x

is an accumulation point of S

23 Leta0 anda1 be distinct real numbers De nean =a n 1 +an 2 for each2 positive integer n 2 Show that fang1n=1 is a Cauchy sequence You may want to use induction to show that

1 n

an+1 an = 2 (a1 a0) and then use the

result from Example 0.9 of Chapter 0

The statement an+1 an = 2 n (a1 a0) is obviously true for n =

1 Suppose that it is true for n < N Then, a a = aN+1 +aN

1 0 N i =

(

a

a a )+

N+1

+aN aN

18

N+1 1 N 1 N N 1

2 1 0 2 1 0 2

1 N i

0 (a a ) 1 0

a ) = + (a a ) = 2 + 1 1 0 Thus, t he s tat ement is proven by i nducti on N 1 2

N+2 N+1 2

1 N+1 a N+1 a N

(a a ) = 1(a Now, let > 0 and n; m be given Choose N = lg ja1 a0 j(n m ) j j jj j

Then,j + a + a2 (a1 a0+ + ) a n 2= +< + a n m m +1for n N: Thus, we see that am < an an 1 + + am+1ja n a m j = am < a n 1 n 1

n m n m n m

24 Suppose fang1n=1 converges to A and fan : n 2 Ng is an in nite set Show that A is an accumulation point of fan : n 2 Ng k for all n N Thus, f n 2 Ng k 1 Let Nk = A k1 ; a + k1 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 in nity of Nk's, we have an in nity of members of fang in any neighborhood 1.3 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 do not converge but fan + bng1n=1 converges 2

(a1 1 1 0

h

+ 1

a0) = h

+ 1NN

1 + 2

N+1i2

0

a )

= 1 2 i

1 (a

l

m

CHAPTER 1 SEQUENCES 19

Let an = ( 1)n and bn = ( 1)n+1 We know that an and bn don't converge, but

an + bn ! 0

27 Suppose fang1n=1 and fbng1n=1 are sequences such that fang1n=1 con-verges to A 6= 0 and fanbng1n=1 Prove that fbng1n=1 converges

.

Suppose anbn ! C Then, fbng = an , so by Theorem 1.9, bn ! A

28 If a 0 for all n, show

1 converges to a with a n n=1 converges

CHAPTER 1 SEQUENCES 18

j j a

Let > 0 be given Then, there is an N such that a n a < p Then, p n p a n a+n p a a j n j pa n+pa j n j pa

a a = = a a 1 < a a 1 < for all n N

8 9 29 Prove that n + k 1

> > > > < k =

> (n + k)k >> > : ; n=1 converges to 1 , where k !

n + k = (n + k)!:

k n!k!

We see that n (n+ !(k n)! + k) k = n (n+ k k!(n 1)( +k) n+k k2) 1 (n+1) o Now, let > 0 be

n!k 1 k o ( n+k 1) k(n+1) 1 (n + k) k 2 (n+1) k 1 1

given Choose N = Then, 1 < =

)

k! k!(n+k ) k! n+1 1 = n+1 n k = 1 k k!(n+k1 k N.

< < for all n

k!(n+ k) k! k!(n+ k) k!(n+ k)

30 Prove the following variation on Lemma 1.10 If fbng1n=1 converges to B 6= 0 and bn 6= 0 for all n, then there is M > 0 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

a + a + + a

n= 1 2 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 > 0 be given There is an N1 such that jan Aj < 2 for all n > N

Let M = ja1 Aj + ja2 Aj + + ja N1 Aj Then, there is an N2 such that M < for all n > N Let N = max(N ; N ) Then,

a1 +a 2 + +a n A = n n 2 2j j n 1 2 n n n

< a 1 + +aN+ +an nA < a A + +jaN Aj + ja N+1 Aj+ +jan Aj = M + ja N+1 Aj+ +jan Aj

2 + 2 = (A quicker way to show this would be to observe that by the de nition

of convergence, a < for all n > N for some N Then, since

j n 2

Let a = ( 1)n Then, as we have seen before, ag diverges, but a 1 +a 2 + +a n 0: n f n

n !

32 Find the limit of the sequences with general term as given:

(a) n2+4 n n2 n5

(b) cos

n

sin n 2

(c) n

CHAPTER 1 SEQUENCES 19

(e) q 4 n p n n1 2 n

33 Find the limit of the sequence in Exercise 23 when a0 = 0 and a1 = 3 You might want to look at Example 0.10 must be 2 2 2n1

2

n1 !

Example 0.10 says that an 1

+a

n 2 = 3 + 2 Since 0, the limit

1.4 Subsequences and Monotone Sequences

34 Find a convergent subsequence of the sequence

1 1

(f) ( 1)

(a) 1

(b) 0

(c) 0

(d) 0

(e) 4 n1 2 n = n 4 n 4nn

Thus, limit is 1

(f) 0

2n =

4 n2

n p 4 n2

= 4 4nn2 n+

n4n p

2

4n 2

p

2

Let nk = 2n Then, the subsequence is 1 2 1n which converges to 0

35 Suppose x is an accumulation point of fan : n 2 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 in nity of fang Thus, let ank be a member of fan : n 2 Ng \ x k1 ; x + k1 Then, for any > 0, there is a K such that k1 < 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 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 =

M Let > 0 be given Then, there are a1 M intervals that sequence values

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 2 (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 pcgn =1 converges to 1

It is clear that n p n c > n+1pc Thus,

f pc n g is a monotone decreasing sequence n pp

Also, c > 1, so by Theorem 1.16, c ! 1

CHAPTER 1 SEQUENCES 21

*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 1 = 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 = 6 and an = 6 + an1 for n > 1 is convergent and nd its limit

p

To nd the limit L, set L = 6 + L , L2 = 6 + L , L2 L 6 = 0 The solutions are 2 and 3

provepthat fan g is decreasing Since 6 + an1 < 6 + an1 and we know that an

1 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 do 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 fxng Thus, sup E; inf E 2 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 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