Chapter 1 Sequences

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1.1 Sequences and Convergence

1. Show that [0; 1] is a neighborhood of 2 { that is, there is > 0 such that

32 ; 3

2 + 1[0; 1].

3

2

1 2

+

1

=

1

[0; 1].

Choose = 3 . Then 3 3 ; 3 3 3 ; 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 2 yj . 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 + ).

Let = min jy (x+ )j

; jy (x )j

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

2 2

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.

13

CHAPTER 1. SEQUENCES 14

6. Use the de nition of convergence to prove that each of the following sequences converges:

(a) 5 + 1 1

(b) n n n=1

f n=1

2 2n 1

gn=1

(c) 2 n 1

(d) n

3n 1

on=1

2n+1 = . Then, n + = n

(b) Let > 0 be given. Let N 2

< < . =

(a) Let > 0 be given. Let N = 1 . Then, 5 + n1 5 = n

1 N1

2 < 2 < .

n N 2 2 2n 2 2n+2n

1

Let 1 n 1

(c) > 0 be given. Let N = lg . Then, j

2 = 2n < 2N < .

> 0 be given. Let N = 3 Then, 3n j 3 = 6n 6n 3 =

(d) Let 4

2n+1

2

4n+2

4n+2 < 4n < 4N < .

3 3 3

*7. Show that fang1n=1 converges to A i fan Ag1

n=1 converges to 0.

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

8. Suppose fang1n=1 converges to A, and de ne a new sequence fbng1

n=1 by bn =

an+an+1 for all n. Prove that fbng1

n=1 converges to A. 2

a A a A N 2 A = 2 = 2

Let > 0 be given. We see that an

+an+1 a

n+a

n+1 2A an A+an+1 A

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

!

A.

2 2 s.t. 2 2

*9. Suppose fang1n=1, fbn g1

n=1, and fcng1n=1 such that fang1

n=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, < an 1 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 fjanjg1

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, 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 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 > 0 be given and set N = x . Then, a x x

x+2x+2 +nx x =

x(1+ 2 +n) x = xn(n2 x = xn2 + xn2 x =

< x n < .

2

n 2 2n 2 2 2 2n n 2 2N

2n 2 2n

+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 fbng1

n=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 fbng1

n=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 < .

CHAPTER 1. SEQUENCES 16

17. Prove that the sequence

2n+1 1

n n=1 is Cauchy.

nm = mn mn < N2 = N < .

2mn+m 2mn n

Let > 0 be given. Choose N = 1 . Then, 2n+1

2m+1 = =

n m

nm

m n m n N 1

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

Let an =

1 n

;;

if n is even

1=

+ 1=n if n is odd . Then, an has accumulation points at

0 and 1.

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 2 Q, the sequence a

b + n1 converges to a

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 + k

1 : n and k are positive integers .

The set f2n : n 2 Z+g[f1g is the set of accumulation points since 2n + k

1 ! 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 n1 < an < x. Since x n

1 ! 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

2 for each

2

positive integer n 2. Show that fang1n=1 is a Cauchy sequence. You may want

to use induction to show that

an+1 an =

1

n

(a1 a0)

2

and then use the result from Example 0.9 of Chapter 0.

CHAPTER 1. SEQUENCES 17

The statement an+1 an = 21

n (a1 a0) is obviously true for n =

1. Suppose that it is true for n < N. Then, a

N+2 a

N+1 = a

N+1+a

N

2

h i

1 N aN+1+aN 2aN 1 N+1 a

N+1 a

N

1

(a a ) =

1 =

(a

a ) =

1

+ (a

(a a )+ 1 + 1

2 (a1 a0) + aN

1

2

+ 2

1 (a1 a0) = 2

N+1 1 N 1 N N h 1 N N N+1 i

1 0

1

2 1 0 2 1

i

0 2

h

2

i

2

1 N

a ).

0

(a

1

0

1

0

a ) =

+

(a

a

) =

+ 1

(a

a

) =

+1

0 N

1

2 Thus, the s tatement is proven by i nducti on.

m

1 n 2 l ja1 a0j(n m)

Now, let > 0 and n; m

be given. Choose N = lg .

n j j jj j

Then,

2

(a1 a0)

<

for n N: Thus, we see that jan amj =

n m

j + + + = .

am < an a

n 1 + +

a

m+1

am <

a a

n 1 +

a

n 1 a

n 2 + + am+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 + k

1 . 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 fang1

n=1 and fbng1n=1 are sequences such that fang1

n=1

and fan + bng1n=1 converge. Prove that fbng1

n=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 fbng1

n=1 do 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 ! 0.

27. Suppose fang1n=1 and fbng1

n=1 are sequences such that fang1n=1

con-verges to A 6= 0 and fanbng1n=1. Prove that fbng1

n=1 converges. n o

Suppose anbn ! C. Then, fbng =

anbn

, so by Theorem 1.9, bn !

C

an A .

28. If a

0 for all n, show

1

converges to a with an

pan

converges tofpna

g.n1=1

n=1

CHAPTER 1. SEQUENCES 18

j j a

p Let > 0 be given. Then, there is an N such that an a < p . Then,

n

pan+

p a j n j pan+pa j n j pa

a =

an a

= a a

1

< a

a

1

< for all n

N.

a

29. Prove that 8

n + k 9

1

> > > >

< k =

> (n + k)k > > >

: ; n=1

converges to k1

! , where

n + k =

(n + k)! :

k

n!k!

We see that

(n+k)! = (n+k 1)(n+k k2)1 (n+1) . Now, let > 0 be

!(n+k) k

o n

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

(n+k)k 2(n+1)

n

n!k1 k 1) (n+1) 1 1

given. Choose N =

. Then,

k 1

<

=

k!

k!(n+k) k 1

k!

1 k k!(n+k )

k

n+1 1 = n+1 n k = 1 N.

<

< fo r all n

k!(n +k) k!

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

n

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

n = a

1 +

a

2 + +

a

n : n

Prove that if fang1n=1 converges to A, then f ng1

n=1 converges to A. Give an

example in which f ng1n=1 converges, but fang1

n=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 + + jaN1 Aj. Then, there is an N2 such that

M < for all n > N . Let N = max(N ; N ). Then, a1+a2+ +an A = n <

n 2 n 2 j j n n n

1 2

n

< a

1

a

1 + +aN + +an nA A + +jaN Aj + jaN+1 Aj+ +jan Aj = M + jaN+1 Aj+ +jan Aj

2 + 2 = . (A quicker way

to show this would be to observe that by the de nition

of convergence, j

a nj 2

for all n > N for some N. Then, since

<

Let a = ( 1)n. Then, as we have seen before,

f a

ng

diverges, but

a1+a

2+ +a

n

!

0: n n

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

(a) n2+4n

n2 n5

(b) cos

n

(c) sin n2

p

n

CHAPTER 1. SEQUENCES 19

(d) n

n2 3

q n pn

(e) 4 1 2 n

n

(f) ( 1)

n+7

(a) 1 (b) 0 (c) 0 (d) 0

q q

(e) 4 n1 2 n = n 4 n

1 4n

n .

Thus, limit is 1

4 . (f) 0

2n =

p

p

=

4n2 n 4n

2

4n2 n 4n2

p 4n n+ 4n

2

2 p

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 2n 2n !

Example 0.10 says that a

n 1+a

n 2 = 1

3 + 2. Since 1

0, the limit

1.4 Subsequences and Monotone Sequences

34. Find a convergent subsequence of the sequence

1 1

( 1)n 1

n=1

n

Let nk = 2n. Then, the subsequence is 1 21n 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 + k

1 . Then,

for any > 0, there is a K such that k1 < for all k > K. Thus, ank ! x.

36. Let fang1

n=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 1. 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 fang1

n=1 con-verges.

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.

n

converges to 1.

38. Prove that if c > 1, then f p

cgn1=1

It is clear that n n+1 n is a monotone decreasing sequence.

pc > p

c. Thus, pc

Also,

n f g

n

pc > 1, so by Theorem 1.16,

pc ! 1.

*39. Suppose fxng1n=1 converges to x0 and fyng1

n=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 + an 1 for n > 1 is convergent and nd its limit.

p

To nd the limit L, set L = 6 + L , L2 = 6 + L , L

2 L 6 = 0. The

solutions are 2 and 3. The only solution that works is 3. Thus, a ! 3. We p p p n

provepthat fan g is decreasing. Since 6 + an 1 < 6 + an 1 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

CHAPTER 1. 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 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)g

1n=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 .

43. Assume 0 a b. Does the sequence f(an + bn)1=ng1

n=1 converge or diverge? If the sequence converges, nd the limit.

The sequence does not converge in general. For instance, if a = 1 and b =

1, then the sequence becomes f[1n + ( 1)n]1=ng. Taking even indexes, the limit

is 1, and taking odd indexes, the limit is 0. Thus, not all subsequences converge to the same limit point, so the sequence is not convergent.

44. Does the sequence

P

( k

p

1

)1

k2 + n

n=1

k=1

X

diverge or converge? If the sequence converges, nd the limit.

k 1 1 1 k 1 1

We see that

n=1 p

<

n=1 k = k = 1. Also, n=1 p

>

k2+n k

2+n

1 p 1 = p k

! 1. Thus, the sequence must converge to 1 by the

n=1

2 k2+k

k +k P P 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 2 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 2 R n Q for all n and xn ! x.

CHAPTER 1. SEQUENCES 22

47. Suppose that fang1n=1 converges to A and that B is an accumulation

point of fan : n 2 Ng. Prove that A = B.

If B is an accumulation point, then B n1 ; B + n

1 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 fbng1

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=bng1

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 2 (L ; L + ) for all n N. Thus, for the same N,

an 2 (bn[L ]; bn[L + ]). Thus, an bn(L + 1). Thus, an is O(bn) if N is

chosen to correspond to = 1. Similarly, bn is O(an).

a 0 and f

b ng

is bounded or b

n ! 1 and a is

If L = 0, then either 2 n ! f ng

n ! 0 and

1=n

bounded. For instance, n3

= 0. (This result is called the Limit

2+1=n

Comparison Test.