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Theorem 0 1 [Cauchy Convergence Criterion] A sequence of real numbers is convergent if and only if it is a Cauchy sequence Proof Let {

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CAUCHY SEQUENCE

DR.HAMED AL-SULAMI

De¯nition 0.1.

A sequencefxngof real numbers is said to beCauchy sequenceif for every" >0 there existsN2Nsuch that ifn;m > N) jxn¡xmj< ": A sequence is Cauchy if the terms eventually get arbitrarily close to each other.

Example 0.1.

The sequencef1

n gis Cauchy. To see this let" >0 be given. ChooseN2Nsuch that 1 N 2 :Now, ifn;m > N) j1 n ¡1 m j ·1 n +1 m 2 2

Example 0.2.

The sequencefn

n+1gis Cauchy. To see this let" >0 be given. ChooseN2Nsuch that 1 N 2 :Now, ifn;m > N) jn n+1¡m m+1j=jm+1¡n¡1 (n+1)(m+1)j · jm¡n nm j<1 n +1 m 2 2

Lemma 1.

Let sequencefxngbe a Cauchy sequence of real numbers. Thenfxngis bounded.

Proof.

Sincefxngis a Cauchy sequence, then there existsN2Nsuch that ifn;m > N) jxn¡xmj<3: ifn;m > N) jxn¡xmj<3 letm=N+ 1, ifn > N) jxn¡xN+1j<3 Note:jxnj ¡ jxN+1j · jxn¡xN+1j ) jxnj ¡ jxN+1j · jxn¡xN+1j<3 ifn > N) jxnj<3 +jxN+1j:

LetM= maxfjx1j;jx2j;¢¢¢jxNj;jxN+1j+ 3g

Now, ifn > N) jxnj<3 +jxN+1j ·M

Now, ifn·N) jxnjThus8n2N;jxnj ·M:

Date: April 1, 2006.

Cauchy Sequence Dr.Hamed Al-Sulami

Theorem 0.1.[Cauchy Convergence Criterion]

A sequence of real numbers is convergent if and only if it is a Cauchy sequence.

Proof.

Letfxngbe a sequence of real numbers.

()) Suppose that limn!1xn=x2R:We want to show thatfxngis Cauchy sequence.

Let" >0 be given. Since limx!1xn=x)9N2N3

ifn > N) jxn¡xj<" 2

Also, ifm > N) jxm¡xj<"

2 Now, ifn;m > N) jxn¡xmj=jxn¡x+x¡xmj · jxn¡xj+jxm¡xj<" 2 2

Thusfxngis a Cauchy sequence:

(() Suppose thatfxngis a Cauchy sequence.We want to show thatfxngis convergent. Let" >0 be given. Sincefxngis a Cauchy sequence, then by Lemma 1 it is bounded. Hencefxnghas a converge subsequencefxnkg:Suppose limk!1xnk=x2R:

There existN1;N22N3ifn;m > N1) jxn¡xmj<"

2 and, ifk > N2) jxnk¡xj<" 2 Now, ¯xk > N2such thatnk> N1and, ifn > N1) jxn¡xnkj<" 2 andjxnk¡xj<" 2 Now, ifn > N1) jxn¡xj=jxn¡xnk+xnk¡xj · jxn¡xnkj+jxnk¡xj<" 2 2

April 1, 2006 2

c°Dr.Hamed Al-Sulami

Cauchy Sequence Dr.Hamed Al-Sulami

Example 0.3.Prove that any sequence of real numbersfxngwhich satis¯es jxn¡xn+1j=1 5 n;8n2Nis convergent.

Proof.

Ifm > n) jxn¡xmj=jxn¡xn+1+xn+1+xn+2+:::+xm¡1¡xmj

· jxn¡xn+1j+jxn+1+xn+2j+:::+jxm¡1¡xmj

1 5 n+1 5 n+1+:::+1 5 m¡1 1 5 n¡1µ 1 5 +1 5

2+:::+1

5 1 5 n¡1m¡nX k=11 5 k 1 5 n¡1µ

1¡1

Note that:µ

1¡1

5 <1 1 5 n¡1:

Let" >0 be given, chooseN2Nsuch that1

5 n¡1< ":

Now, ifn;m > N) jxn¡xmj<1

5 n¡1< ":

April 1, 2006 3

c°Dr.Hamed Al-Sulamiquotesdbs_dbs6.pdfusesText_11

[PDF] [PDF] CAUCHY SEQUENCE Definition 01 A sequence {xn} of real

CAUCHY SEQUENCE

DR.HAMED AL-SULAMI

De¯nition 0.1.

Example 0.1.

The sequencef1

Example 0.2.

The sequencefn

Lemma 1.

Proof.

LetM= maxfjx1j;jx2j;¢¢¢jxNj;jxN+1j+ 3g

Now, ifn > N) jxnj<3 +jxN+1j ·M

Date: April 1, 2006.

Cauchy Sequence Dr.Hamed Al-Sulami

Theorem 0.1.[Cauchy Convergence Criterion]

Proof.

Letfxngbe a sequence of real numbers.

Let" >0 be given. Since limx!1xn=x)9N2N3

Also, ifm > N) jxm¡xj<"

Thusfxngis a Cauchy sequence:

There existN1;N22N3ifn;m > N1) jxn¡xmj<"

April 1, 2006 2

Cauchy Sequence Dr.Hamed Al-Sulami

Proof.

· jxn¡xn+1j+jxn+1+xn+2j+:::+jxm¡1¡xmj

2+:::+1

1¡1

Note that:µ

1¡1

Let" >0 be given, chooseN2Nsuch that1

Now, ifn;m > N) jxn¡xmj<1

April 1, 2006 3