3.11: Operations on Convergent Sequences
This page is a draft and is under active development.
( \newcommand{\kernel}{\mathrm{null}\,}\)
Sequences in E1 and C can be added and multiplied termwise; for example, adding {xm} and {ym}, one obtains the sequence with general term xm+ym. This leads to important theorems, valid also for En(* and other normed spaces). Theorem 1 below states, roughly, that the limit of the sum {xm+ym} equals the sum of lim xm and lim ym (if these exist), and similarly for products and quotients (when they are defined).
Let xm→q,ym→r, and am→a in E1 or C (the complex field). Then
(i) xm±ym→q±r;
(ii) amxm→aq;
(iii) xmam→qa if a≠0 and for all m≥1,am≠0.
This also holds if the xm,ym,q, and r are vectors in En ("or in another normed space), while the am and a are scalars for that space.
- Proof
-
(i) By formula (2) of §14, we must show that
(∀ε>0)(∃k)(∀m>k)|xm±ym−(q±r)|<ε.
Thus we fix an arbitrary ε>0 and look for a suitable k. since xm→q and ym→r, there are k′ and k′′ such that
(∀m>k′)|xm−q|<ε2
and
(∀m>k′′)|ym−r|<ε2
(as ε is arbitrary, we may as well replace it by 12ε). Then both inequalities hold for m>k,k=max(k′,k′′). Adding them, we obtain
(∀m>k)|xm−q|+|ym−r|<ε.
Hence by the triangle law,
|xm−q±(ym−r)|<ε, i.e., |xm±ym−(q±r)|<ε for m>k,
as required. ◻
This proof of (i) applies to sequences of vectors as well, without any change.
The proof of (ii) and (iii) is sketched in Problems 1-4 below.
Note 1. By induction, parts (i) and (ii) hold for sums and products of any finite (but fixed) number of suitable convergent sequences.
Note 2. The theorem does not apply to infinite limits q,r,a.
Note 3. The assumption a≠0 in Theorem 1( iii) is important. It ensures not only that q/a is defined but also that at most finitely many am can vanish (see Problem 3). Since we may safely drop a finite number of terms (see Note 2 in §14), we can achieve that noam is 0, so that xm/am is defined. It is with this understanding that part (iii) of the theorem has been formulated. The next two theorems are actually special cases of more general propositions to be proved in Chapter 4, §§3 and 5. Therefore, we only state them here, leaving the proofs as exercises, with some hints provided.
(componentwise convergence). We have ¯xm→¯p in En(∗Cn) iff each of the n components of ¯xm tends to the corresponding component of ¯p, i.e., iff xmk→pk,k=1,2,…,n, in E1(C). (See Problem 8 for hints.)
Every monotone sequence {xn}⊆E∗ has a finite or infinite limit, which equals sup_ nxn if {xn}↑ and inf nxn if {xn}↓. If {xn} is monotone and bounded in E1, its limit is finite (by Corollary 1 of Chapter 2, §13).
The proof was requested in Problem 9 of Chapter 2, §13. See also Chapter 4, §5, Theorem 1. An important application is the following.
(the number e).
Let xn=(1+1n)n in E1. By the binomial theorem,
xn=1+1+n(n−1)2!n2+n(n−1)(n−2)3!n3+⋯+n(n−1)⋯(n−(n−1))n!nn=2+(1−1n)12!+(1−1n)(1−2n)13!+⋯+(1−1n)(1−2n)⋯(1−n−1n)1n!
If n is replaced by n+1, all terms in this expansion increase, as does their number. Thus xn<xn+1, i.e., {xn}↑. Moreover, for n>1,
2<xn<2+12!+⋯+1n!≤2+12+⋯+12n−1=2+12(1+⋯+12n−2)=2+121−(12)n−112<2+1=3
Thus 2<xn<3 for n>1. Hence 2<supnxn≤3; and by Theorem 3, supnxn=limxn. This limit, denoted by e, plays an important role in analysis. It can be shown that it is irrational, and (to within 10−20) e=2.71828182845904523536… In any case,
2<e=limn→∞(1+1n)n≤3.
The following corollaries are left as exercises for the reader.
Suppose limxm=p and limym=q exist in E∗.
(a) If p>q, then xm>ym for all but finitely many m.
(b) If xm≤ym for infinitely many m, then p≤q; i.e., limxm≤limym.
This is known as passage to the limit in inequalities. Caution: The strict inequalities xm<ym do not imply p<q but only p≤q. For example, let
xm=1m and ym=0.
Then
(∀m)xm>ym;
yet limxm=limym=0.
Let xm→p in E∗, and let c∈E∗ (finite or not). Then the following are true:
(a) If p>c (respectively,p<c), we have xm>c(xm<c) for all but finitely many m.
(b) If xm≤c (respectively, xm≥c) for infinitely many m, then p≤c (p≥c).
One can prove this from Corollary 1, with ym=c (or xm=c) for all m.
(rule of intermediate sequence). If xm→p and ym→p in E∗ and if xm≤zm≤ym for all but finitely many m, then also zm→p.
(continuity of the distance function). If
xm→p and ym→q in a metric space (S,ρ),
then
ρ(xm,ym)→ρ(p,q) in E1.
- Proof
-
Hint: Show that
|ρ(xm,ym)−ρ(p,q)|≤ρ(xm,p)+ρ(q,ym)→0
by Theorem 1.