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(3 3) Theorem A sequence (xn) of real numbers converges if and only if it is a Cauchy sequence Proof: Assume that (xn) converges to a real number XO' Let the

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[PDF] (32) Definition A sequence (xn) of real numbers is a Cauchy

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3. Sequences and Series of Real Numbers 29

(3.2) Definition. A sequence (xn) of real numbers is a Cauchy sequence if for each k in Z+ there exists Mk in Z+ such that (3.2.1) (3.3) Theorem. A sequence (xn) of real numbers converges if and only if it is a Cauchy sequence.

Proof:

Assume that (xn) converges to a real number XO' Let the sequence (N k) satisfy (3.1.1). Write Mk==N zk ' Then

IXm-xnl IX

m -xol +Ixn-xol +(2k}-1 =k-

1 for m, Therefore (xn) is a Cauchy sequence.

Assume conversely

that (xn) is a Cauchy sequence. Let the se quence (M k) satisfy (3.2.1). Write Nk==max{3k,M 2k }. Then

IXm-xnl Nk).

Let Yk be the (2kth rational approximation to X

Nk • For

IYm -Ynl :s; IYm -XNJ + IXNm -XNJ + IXNn -Ynl

:s;(2m}-1 +(2m}-1 +(2n)-1 +(2n}-1 =m- 1 +n- 1.

Therefore

Y==(Yn) is a real number. To see that (xn) converges to y, we consider n Nk and compute

Iy-xnl Ynl +IYn-XNJ +IXNn -xnl

n- 1 +(2n}-1 +(2k}-1 + (6k}-1 +(2k}-1 = k- 1. D A subsequence of a convergent sequence converges to the same limit. If a sequence converges, then any sequence obtained from it by modifications (including, perhaps, insertions or deletions) which in volve only a finite number of terms converges to the same limit. If x == (xn) is a regular sequence of rational numbers, then (x:) converges to x, by (2.14).

A sequence (xn) is increasing

(respectively, strictly increasing) if xn+ 1 Xn (respectively, xn+ 1> Xn) for each n. Decreasing and strictly decreasing sequences are defined analogously, in the obvious way. A theorem of classical mathematics states that every bounded increasing sequence of real numbers converges. A counterexample to this state ment is given by any increasing sequence (xn) such that Xn = 0 or xn = 1 for each n, but it is not known whether xn = 0 for all n. It is useful to supplement Definition (3.1) by writing limxn= 00 or

30 Chapter 2 Calculus and the Real Numbers

to express the fact that for each k in Z+ there exists Nk in Z+ with xn>k for all n?;,Nk. We also define limxn= -00 or to mean that lim -Xn = 00. The next proposition shows that we may work with real numbers constructed as limits by working with their approximations. (3.4) Proposition. Assume that xn-+xO as n-+oo, and Yn-+Yo as n-+oo, where Xo and Yo are real numbers. Then (a) xn+ Yn-+xO+ Yo as n-+oo (b) xnyn-+xoYo as n-+oo (c) max{xn,Yn}-+max{xo,Yo} as n-+oo (d)

Xo = c whenever Xn = c for all n

(e) if xo=l=O and xn=l=O for all n, then X;l-+xol as n-+oo (f) if for all n, then Proof: (a) For each k in Z+ there exists Nk in Z+ such that Then

Therefore xn+yn-+xO+Yo as n-+oo.

(b) Choose m in Z+ such that IYol and Ixnl for all n. For each k in Z+ choose Nk in Z+ with IX n -xol k)-\ IYn -Yol k)-l (n?;, Nk).

Then for n?;, Nk,

IXn Yn -Xo Yol IXn(Yn -Yo)1 + IYo(xn -xo)1

Yol

Therefore xnyn-+xoYo as n-+oo.

Imax{x

Yn} -max{xo, Yo}1 IYn-Yol},

it follows that

3. Sequences and Series of Real Numbers 31

(d) If Xn = c for all n, then (xn) converges to c. Therefore Xo = c. (e) Since IXol >0,

IXnl IXol-lxn -xol >i: IXol

whenever n is large enough, say for n no. Let k and n be positive integers such that and IXn-xol "2k)-1IxoI2. Then

Ix; I_Xo 11 = IXnl-llxol-llxn -xol =k-

Therefore

x; 1-+ Xo 1 as n -+ 00. (f) We compute Yo -Xo = lim Yn -lim Xn = lim (Yn -xn) = lim IYn -xnl = lim max{Yn-xn, xn-Yn} =max{yo-xo,xo-Yo} by (a), (b), (c), and (d). D

For each sequence (xn) of real numbers the number

is called the nIh partial sum of (xn), and (sn) is called the sequence of partial sums of the sequence (xn). A sum So of (xn) is a limit of the sequence (sn) of partial sums. We write to indicate that So is a sum of (xJ A sequence which is meant to be summed is called a series. A series is said to converge to its sum. Thus the sequence (2 -n}:,= 1 converges to 0 as a sequence, but as a series it 00 converges to L 2 -n = l. n=1 A convergent series remains convergent, but not necessarily to the same sum, after modification of finitely many of its terms. 00 The series (xn) is often loosely referred to as the series L xn.

00 n= 1

If the series L xn converges, then xn-+O as n-+oo.

00 n= 1 00

A series L Xn is said to converge absolutely when the series L IXnl converges. n= 1 n= 1 In classical analysis a series of nonnegative terms converges if the partial sums are bounded. This is not true in constructive analysis.

However,

we have the following result.

32 Chapter 2 Calculus and the Real Numbers

00 (3.5) Proposition. If I Y n is a convergent series of nonnegative terms, n= 1 00 and if IXnl Yn for each n, then I Xn converges. n= 1 00

Proof: Since I Y

n is convergent, the sequence of partial sums is a n= 1 Cauchy sequence. Therefore for each k in 7l+ there exists an Nk in 7l+ with m I Then j=n+ 1 00 Therefore the sequence of partial sums of the series I Xn is a Cauchy sequence.

By (3.3), the series converges. D n= 1

The criterion of Proposition (3.5) is known as the comparison test. It follows from the comparison test that every absolutely convergent series is convergent. 00 The terms of an absolutely convergent series I Xn may be re- n= 1 ordered without affecting the sum So of the series. More precisely, if A: 00

7l+ -+71+ is a bijection, then I Xl(n) exists and equals so. This may

00 n= 1

not be true if the series I Xn is merely convergent. n= 1 A sequence (xn) is said to diverge if there exists E in IR. + such that for each k in 7l+ there exist m and n in 7l+ with m, and IX m -xnl E. The motivation for this definition is, of course, that a se quence cannot be both convergent and divergent. A series is said to diverge if the sequence of its partial sums diverges. 00

The series I n -1 diverges, because

n= 1 00 The series I Xn diverges whenever there exists r in IR. + such that n= 1

IXnl r for infinitely many values of n.

00 00

Let I Xn and I Y

n be series of nonnegative terms. The compari- n=1 n=1 00 00 son test for divergence is that I Xn diverges whenever I Y n diverges n=1 n=1 and there is a positive integer N with xn Y n for all n N.

3. Sequences and Series of Real Numbers 33

The following very useful test for convergence and divergence is called the ratio test. 00 (3.6) Proposition. Let L Xn be a series, c a positive number, and N a n= 1 00 positive integer. Then L Xn converges if c < 1 and n= 1 (3.6.1 ) and diverges if c> 1 and (3.6.2) Proof: Assume that c< 1 and that (3.6.1) is valid. Then Ixnl :::;;cn-NlxNI 00 for n N. By the comparison test, L xn converges. n= 1 Next, assume that c> 1 and that (3.6.2) holds. Then

IXnl 11 IxN+ 11 N + 1)

and 00

Therefore I Xn diverges. 0

n= 1 A corollary of the ratio test is that if the limit

L= limlxn+lx;11

00 exists, then L Xn converges whenever L< 1 and diverges whenever

L>1. n=1

The ratio test says nothing in case L = 1. To handle this case, we introduce stronger tests based on Kummer's criterion. (3.7) Lemma. Let (an) and (xn) be sequences of positive numbers, c a 00 positive number, and N a positive integer. Then L xn converges if anxn--+O as n--+oo and n=l (3.7.1) 00 00 while L xn diverges if I a; 1 diverges andquotesdbs_dbs6.pdfusesText_11

[PDF] [PDF] (32) Definition A sequence (xn) of real numbers is a Cauchy

3. Sequences and Series of Real Numbers 29

Proof:

IXm-xnl IX

1 for m, Therefore (xn) is a Cauchy sequence.

Assume conversely

IXm-xnl Nk).

Let Yk be the (2kth rational approximation to X

IYm -Ynl :s; IYm -XNJ + IXNm -XNJ + IXNn -Ynl

Therefore

Iy-xnl Ynl +IYn-XNJ +IXNn -xnl

A sequence (xn) is increasing

30 Chapter 2 Calculus and the Real Numbers

Xo = c whenever Xn = c for all n

Therefore xn+yn-+xO+Yo as n-+oo.

Then for n?;, Nk,

IXn Yn -Xo Yol IXn(Yn -Yo)1 + IYo(xn -xo)1

Therefore xnyn-+xoYo as n-+oo.

Imax{x

Yn} -max{xo, Yo}1 IYn-Yol},

3. Sequences and Series of Real Numbers 31

IXnl IXol-lxn -xol >i: IXol

Ix; I_Xo 11 = IXnl-llxol-llxn -xol =k-

Therefore

For each sequence (xn) of real numbers the number

00 n= 1

00 n= 1 00

However,

32 Chapter 2 Calculus and the Real Numbers

Proof: Since I Y

By (3.3), the series converges. D n= 1

7l+ -+71+ is a bijection, then I Xl(n) exists and equals so. This may

00 n= 1

The series I n -1 diverges, because

IXnl r for infinitely many values of n.

Let I Xn and I Y

3. Sequences and Series of Real Numbers 33

IXnl 11 IxN+ 11 N + 1)

Therefore I Xn diverges. 0

L= limlxn+lx;11

L>1. n=1