SEQUENCES AND SERIES
Section 3.6 Properly Divergent Sequences
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 ðxnynÞare Cauchy sequences.
5. Ifxn:¼pffiffiffin, show that (xn) 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 (xn) is convergent. What is its limit?
11. If y1<y2 are arbitrary real numbers and yn:¼13yn1þ23yn2forn>2, show that (yn) is convergent. What is its limit?
12. Ifx1>0 andxnþ1:¼ ð2þxnÞ1forn1, show that (xn) 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.
3.6.1 Definition LetðxnÞbe a sequence of real numbers.
(i) We say thatðxnÞtends to1, and write limðxnÞ ¼ þ1, if for everya2R there exists a natural number KðaÞsuch that ifnKðaÞ, thenxn>a.
(ii) We say that (xn)tends to1, and write limðxnÞ ¼ 1, if for everyb2R there exists a natural number K(b) such that ifnKðbÞ, thenxn<b.
We say that ðxnÞis properly divergent in case we have either limðxnÞ ¼ þ1 or limðxnÞ ¼ 1.
The reader should realize that we are using the symbols þ1 and1 purely as a convenient notation in the above expressions. Results that have been proved in earlier sections for conventional limits limðxnÞ ¼LðforL2RÞ may not remain true when limðxnÞ ¼ 1.
3.6.2 Examples (a) limðnÞ ¼ þ1.
In fact, ifa2R is given, letK(a) be any natural number such that K(a)>a.
(b) limðn2Þ ¼ þ1.
If K(a) is a natural number such that K(a) > a, and if nKðaÞ then we have n2n>a.
(c) Ifc>1, then limðcnÞ ¼ þ1.
Letc¼1þb, whereb>0. Ifa2Ris given, letK(a) be a natural number such that KðaÞ>a=b. IfnKðaÞit follows from Bernoulli’s Inequality that
cn¼ ð1þbÞn1þnb>1þa>a:
Therefore limðcnÞ ¼ þ1. &
Monotone sequences are particularly simple in regard to their convergence. We have seen in the Monotone Convergence Theorem 3.3.2 that a monotone sequence is convergent if and only if it is bounded. The next result is a reformulation of that result.
3.6.3 Theorem A monotone sequence of real numbers is properly divergent if and only if it is unbounded.
(a) IfðxnÞis an unbounded increasing sequence, then limðxnÞ ¼ þ1. (b) IfðxnÞis an unbounded decreasing sequence, thenlimðxnÞ ¼ 1.
Proof. (a) Suppose that (xn) is an increasing sequence. We know that if (xn) is bounded, then it is convergent. If (xn) is unbounded, then for anya2Rthere existsnðaÞ 2Nsuch thata<xnðaÞ. But since (xn) is increasing, we havea<xn for allnnðaÞ. Sinceais arbitrary, it follows that limðxnÞ ¼ þ1.
Part (b) is proved in a similar fashion. Q.E.D.
The following ‘‘comparison theorem’’ is frequently used in showing that a sequence is properly divergent. [In fact, we implicitly used it in Example 3.6.2(c).]
3.6.4 Theorem Let(xn)and (yn)be two sequences of real numbers and suppose that
(1) xnyn for all n2N:
(a) IflimðxnÞ ¼ þ1,then limðynÞ ¼ þ1. (b) IflimðynÞ ¼ 1,thenlimðxnÞ ¼ 1.
Proof. (a) If limðxnÞ ¼ þ1, and ifa2R is given, then there exists a natural number K(a) such that if nKðaÞ, then a<xn. In view of (1), it follows that a<yn for all nKðaÞ. Since ais arbitrary, it follows that limðynÞ ¼ þ1.
The proof of (b) is similar. Q.E.D.
Remarks (a) Theorem 3.6.4 remains true if condition (1) is ultimately true; that is, if there existsm2N such thatxnyn for allnm.
(b) If condition (1) of Theorem 3.6.4 holds and if limðynÞ ¼ þ1, it doesnotfollow that limðxnÞ ¼ þ1. Similarly, if (1) holds and if limðxnÞ ¼ 1, it does not follow that limðynÞ ¼ 1. In using Theorem 3.6.4 to show that a sequence tends toþ1[respectively, 1] we need to show that the terms of the sequence are ultimately greater [respectively, less] than or equal to the corresponding terms of a sequence that is known to tend toþ1 [respectively,1].
Since it is sometimes difficult to establish an inequality such as (1), the following
‘‘limit comparison theorem’’ is often more convenient to use than Theorem 3.6.4.
3.6.5 Theorem Let(xn)and(yn)be two sequences of positive real numbers and suppose that for some L2R;L>0,we have
(2) limðxn=ynÞ ¼L:
ThenlimðxnÞ ¼ þ1if and only iflimðynÞ ¼ þ1. Proof. If (2) holds, there existsK2Nsuch that
1
2L<xn=yn<32L for all nK:
Hence we have 12L
yn<xn< 32L
ynfor allnK. The conclusion now follows from a slight modification of Theorem 3.6.4. We leave the details to the reader. Q.E.D.
The reader can show that the conclusion need not hold if eitherL¼0 orL¼ þ1. However, there are some partial results that can be established in these cases, as will be seen in the exercises.
Exercises for Section 3.6
1. Show that if (xn) is an unbounded sequence, then there exists a properly divergent subsequence.
2. Give examples of properly divergent sequences (xn) and (yn) withyn6¼0 for alln2Nsuch that:
(a) ðxn=ynÞis convergent, (b) ðxn=ynÞis properly divergent.
3. Show that ifxn>0 for alln2N, then limðxnÞ ¼0 if and only if limð1=xnÞ ¼ þ1.
4. Establish the proper divergence of the following sequences.
(a) ðpffiffiffinÞ; (b) ð ffiffiffiffiffiffiffiffiffiffiffi
nþ1 p Þ;
(c) ð ffiffiffiffiffiffiffiffiffiffiffi n1
p Þ; (d) ðn= ffiffiffiffiffiffiffiffiffiffiffi
nþ1 p Þ:
5. Is the sequence (nsinn) properly divergent?
6. Let (xn) be properly divergent and let (yn) be such that limðxnynÞbelongs toR. Show that (yn) converges to 0.
7. Let (xn) and (yn) be sequences of positive numbers such that limðxn=ynÞ ¼0.
(a) Show that if limðxnÞ ¼ þ1, then limðynÞ ¼ þ1.
(b) Show that if (yn) is bounded, then limðxnÞ ¼0.
3.6 PROPERLY DIVERGENT SEQUENCES 93
8. Investigate the convergence or the divergence of the following sequences:
(a) ð ffiffiffiffiffiffiffiffiffiffiffiffiffi n2þ2
p Þ; (b) ð ffiffiffi
pn
=ðn2þ1ÞÞ;
(c) ð ffiffiffiffiffiffiffiffiffiffiffiffiffi n2þ1 p = ffiffiffi
pn
Þ; (d) ðsin ffiffiffi
pn Þ:
9. LetðxnÞandðynÞbe sequences of positive numbers such that limðxn=ynÞ ¼ þ1, (a) Show that if limðynÞ ¼ þ1, then limðxnÞ ¼ þ1.
(b) Show that if (xn) is bounded, then limðynÞ ¼0.
10. Show that if limðan=nÞ ¼L, whereL>0, then limðanÞ ¼ þ1.