THE CAUCHY CRITERION 87

SEQUENCES AND SERIES

Section 3.5 The Cauchy Criterion

3.5 THE CAUCHY CRITERION 87

SinceXconverges tox, so does the subsequenceX⁰with odd indices. By Induction, the reader can establish that [see 1.2.4(f)]

x2nþ1 ¼ 1þ1 2þ 1

2³þ þ 1 2²ⁿ¹

¼ 1þ2 3 1 1

4ⁿ

:

It follows from this (how?) thatx¼limX¼limX⁰¼1þ²₃¼⁵₃. (b) Let Y¼ ðy_nÞbe the sequence of real numbers given by

y1 :¼ 1

1!; y2:¼ 1 1!1

2!;. . .; y_n:¼ 1 1!1

2!þ þð1Þ^nþ1 n! ; Clearly,Yis not a monotone sequence. However, if m>n, then

y_my_n¼ð1Þ^nþ2

ðnþ1Þ!þð1Þ^nþ3

ðnþ2Þ!þ þð1Þ^mþ1 m! : Since 2^r1r! [see 1.2.4(e)], it follows that ifm>n, then (why?)

jy_my_nj 1

ðnþ1Þ!þ 1

ðnþ2Þ!þ þ 1 m! 1

2ⁿþ 1

2^nþ1þ þ 1 2^m1< 1

2ⁿ¹:

Therefore, it follows that (yn) is a Cauchy sequence. Hence it converges to a limity. At the present moment we cannot evaluateydirectly; however, passing to the limit (with respect tom) in the above inequality, we obtain

jy_nyj 1=2ⁿ¹:

Hence we can calculate y to any desired accuracy by calculating the terms y_n for sufficiently large n. The reader should do this and show that yis approximately equal to 0.632 120 559. (The exact value ofyis 11=e.)

(c) The sequence 1 1þ1

2þ þ1 n

diverges.

LetH:¼ ðh_nÞbe the sequence defined by

hn:¼1 1þ1

2þ þ1

n for n2N;

which was considered in 3.3.3(b). Ifm>n, then hmhn¼ 1

nþ1þ þ1 m:

Since each of these mn terms exceeds 1=m, then hmhn>ðmnÞ=m¼1n=m. In particular, ifm¼2n we haveh2nhn>¹₂. This shows thatHis not a Cauchy sequence (why?); thereforeHisnota convergent sequence. (In terms that will be introduced in Section 3.7, we have just proved that the ‘‘harmonic series’’X¹

n¼1

1=nis divergent.) &

3.5.7 Definition We say that a sequenceX¼ ðx_nÞof real numbers iscontractiveif there exists a constantC; 0<C<1, such that

jxnþ2xnþ1j Cjxnþ1xnj

for alln2N. The numberCis called theconstantof the contractive sequence.

3.5.8 Theorem Every contractive sequence is a Cauchy sequence, and therefore is convergent.

Proof. If we successively apply the defining condition for a contractive sequence, we can work our way back to the beginning of the sequence as follows:

jx_nþ2x_nþ1j Cjx_nþ1x_nj C²jx_nx_n1j C³jx_n1x_n2j Cⁿjx2x1j:

Form>n, we estimatejx_mx_njby first applying the Triangle Inequality and then using the formula for the sum of a geometric progression (see 1.2.4(f)). This gives

jx_mx_nj jx_mx_m1j þ jx_m1x_m2j þ þ jx_nþ1x_nj

C^m2þC^m3þ þCⁿ¹ jx2x1j

¼ Cⁿ¹ 1C^mn 1C

jx2x1j Cⁿ¹ 1

1C

jx2x1j:

Since 0<C<1, we know limðCⁿÞ ¼0 [see 3.1.11(b)]. Therefore, we infer that (x_n) is a Cauchy sequence. It now follows from the Cauchy Convergence Criterion 3.5.5 that (x_n) is

a convergent sequence. Q.E.D.

3.5.9 Example We consider the sequence of Fibonacci fractionsx_n:¼f_n=f_nþ1 where f1¼f2¼1 and f_nþ1 ¼f_nþf_n1. (See Example 3.1.2(d).) The first few terms are x1¼1;x2 ¼1=2;x3¼2=3;x4¼3=5; x5¼5=8, and so on. It is shown that the se- quence ðx_nÞis given inductively by the equation x_nþ1¼1=ð1þx_nÞas follows:

x_nþ1 ¼f_nþ1

f_nþ2

¼ f_nþ1

f_nþ1þf_n¼ 1 1þ f_n

f_nþ1

¼ 1 1þx_n:

An induction argument establishes 1=2xn1 for alln, so that adding 1 and taking reciprocals gives us the inequality 1=21=ð1þxnÞ 2=3 for alln. It then follows that

jxnþ1xnj ¼ jxnxn1j

ð1þx_nÞð1þx_n1Þ2 32

3jxnxn1j ¼4

9jxnxn1j:

Hence, the sequence (xn) is contractive and therefore converges by Theorem 3.5.8. Passing to the limitx¼limðx_nÞ, we obtain the equationx¼1=ð1þxÞ, so that x satisfies the equation x²þx1¼0. The quadratic formula gives us the positive solution x¼ ð1þ ffiffiffi

p5

Þ=2¼0:618034:. . . The reciprocal 1=x¼ ð1þ ffiffiffi

p5

Þ=2¼1:618034. . .is often denoted by the Greek letterw and referred to as the Golden Ratio in the history of geometry. In the artistic theory of the ancient Greek philosophers, a rectangle havingwas the ratio of the longer side to the shorter side is the rectangle most pleasing to the eye. The number also has many interesting mathematical properties. (A historical discussion of the Golden Ratio can be found on Wikipedia.) &

In the process of calculating the limit of a contractive sequence, it is often very important to have an estimate of the error at thenth stage. In the next result we give two 3.5 THE CAUCHY CRITERION 89

such estimates: the first one involves the first two terms in the sequence andn; the second one involves the differencexnxn1.

3.5.10 Corollary If X:¼ ðxnÞis a contractive sequence with constant C;0<C<1, and if x:¼limX,then

(i) jxxnj Cⁿ¹

1Cjx2x1j;

(ii) jxxnj C

1Cjxnxn1j:

Proof. From the preceding proof, ifm>n, thenjxmxnj

Cⁿ¹=ð1CÞ jx2x1j. If we let m! 1in this inequality, we obtain (i).

To prove (ii), recall that ifm>n, then

jxmxnj jxmxm1j þ þ jxnþ1xnj:

Since it is readily established, using Induction, that

jxnþkxnþk1j C^kjxnxn1j;

we infer that

jx_mx_nj ðC^mnþ þC²þCÞjx_nx_n1j

C

1Cjxnxn1j

We now letm! 1in this inequality to obtain assertion (ii). Q.E.D.

3.5.11 Example We are told that the cubic equation x³7xþ2¼0 has a solution between 0 and 1 and we wish to approximate this solution. This can be accomplished by means of an iteration procedure as follows. We first rewrite the equation asx¼ ðx³þ2Þ=7 and use this to define a sequence. We assign tox1an arbitrary value between 0 and 1, and then define

x_nþ1:¼¹₇ðx³_nþ2Þ for n2N:

Because 0<x1<1, it follows that 0<x_n<1 for all n2N. (Why?) Moreover, we have

x_nþ2x_nþ1 ¼¹₇ðx³_nþ1þ2Þ ¹₇ðx³_nþ2Þ¼¹₇x³_nþ1x³_n

¼¹₇x²_nþ1þxnþ1xnþx²_nxnþ1xn³₇xnþ1xn:

Therefore, (x_n) is a contractive sequence and hence there existsrsuch that limðx_nÞ ¼r. If we pass to the limit on both sides of the equality x_nþ1¼ ðx³_nþ2Þ=7, we obtain r¼ ðr³þ2Þ=7 and hencer³7rþ2¼0. Thus ris a solution of the equation.

We can approximate rby choosing x1, and calculating x2;x3;. . . successively. For example, if we takex1¼0:5, we obtain (to nine decimal places):

x2¼0:303 571 429; x3¼0:289 710 830; x4¼0:289 188 016; x5¼0:289 169 244; x6¼0:289 168 571; etc:

To estimate the accuracy, we note thatjx2x1j<0:2. Thus, afternsteps it follows from Corollary 3.5.10(i) that we are sure thatjxxnj 3ⁿ¹ð7ⁿ²20Þ. Thus, whenn¼6, we are sure that

jxx6j 3⁵=ð7⁴20Þ ¼243=48 020<0:0051:

Actually the approximation is substantially better than this. In fact, since jx6x5j<

0:000 0005, it follows from 3.5.10(ii) thatjxx6j ³₄jx6x5j<0:000 0004. Hence the

first five decimal places ofx6are correct. &

Exercises for Section 3.5

1. Give an example of a bounded sequence that is not a Cauchy sequence.

2. Show directly from the definition that the following are Cauchy sequences.

(a) nþ1 n

; (b) 1þ1

2!þ þ 1 n!

. 3. Show directly from the definition that the following are not Cauchy sequences.

(a)

ð1Þⁿ ; (b) nþð1Þⁿ n

; (c) (lnn)

4. Show directly from the definition that if (xn) and (yn) are Cauchy sequences, then (xnþyn) and ðxny_nÞare Cauchy sequences.

5. Ifxn:¼pffiffiffin, show that (x_n) satisfies limjxnþ1xnj ¼0, but that it is not a Cauchy sequence.

6. Letpbe a given natural number. Give an example of a sequence (xn) that is not a Cauchy sequence, but that satisfies limjxnþpxnj ¼0.

7. Let (xn) be a Cauchy sequence such thatxnis an integer for everyn2N. Show that (xn) is ultimately constant.

8. Show directly that a bounded, monotone increasing sequence is a Cauchy sequence.

9. If 0<r<1 andjxnþ1xnj<rⁿfor alln2N, show that (xn) is a Cauchy sequence.

10. Ifx1<x2 are arbitrary real numbers andxn:¼¹₂ðxn2þxn1Þforn>2, show that (x_n) is convergent. What is its limit?

11. If y1<y2 are arbitrary real numbers and y_n:¼¹₃y_n1þ²₃y_n2forn>2, show that (y_n) is convergent. What is its limit?

12. Ifx1>0 andxnþ1:¼ ð2þxnÞ¹forn1, show that (x_n) is a contractive sequence. Find the limit.

13. Ifx1:¼2 andxnþ1:¼2þ1=xnforn1, show that (xn) is a contractive sequence. What is its limit?

14. The polynomial equation x³5xþ1¼0 has a rootr with 0<r<1. Use an appropriate contractive sequence to calculaterwithin 10⁴.