# 3.11: Operations on Convergent Sequences - Mathematics

Sequences in (E^{1}) and (C) can be added and multiplied termwise; for example, adding (left{x_{m} ight}) and (left{y_{m} ight},) one obtains the sequence with general term (x_{m}+y_{m}). Theorem 1 below states, roughly, that the limit of the sum (left{x_{m}+y_{m} ight}) equals the sum of lim (x_{m}) and lim (y_{m}) (if these exist), and similarly for products and quotients (when they are defined).

Theorem (PageIndex{1})

Let (x_{m} ightarrow q, y_{m} ightarrow r,) and (a_{m} ightarrow a) in (E^{1}) or (C) (the complex field() .) Then

(i) (x_{m} pm y_{m} ightarrow q pm r);

(ii) (a_{m} x_{m} ightarrow a q);

(iii) (frac{x_{m}}{a_{m}} ightarrow frac{q}{a}) if (a eq 0) and for all (m geq 1, a_{m} eq 0).

This also holds if the (x_{m}, y_{m}, q,) and (r) are vectors in (E^{n}) ("or in another normed space(),) while the (a_{m}) and a are scalars for that space.

Proof

(i) By formula (2) of §14, we must show that

[(forall varepsilon>0)(exists k)(forall m>k) quadleft|x_{m} pm y_{m}-(q pm r) ight|

Thus we fix an arbitrary (varepsilon>0) and look for a suitable (k .) since (x_{m} ightarrow q) and (y_{m} ightarrow r,) there are (k^{prime}) and (k^{prime prime}) such that

and

[left(forall m>k^{prime prime} ight) quadleft|y_{m}-r ight|

(as (varepsilon) is arbitrary, we may as well replace it by (frac{1}{2} varepsilon ) .) Then both inequalities hold for (m>k, k=max left(k^{prime}, k^{prime prime} ight) .) Adding them, we obtain

Hence by the triangle law,

[left|x_{m}-q pmleft(y_{m}-r ight) ight|k,]

as required. (square)

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 eq 0) in Theorem 1(() iii) is important. It ensures not only that (q / a) is defined but also that at most finitely many (a_{m}) can vanish (see Problem 3). Since we may safely drop a finite number of terms (see Note 2 in §14), we can achieve that no(a_{m}) is (0,) so that (x_{m} / a_{m}) 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.

Theorem (PageIndex{2})

(componentwise convergence). We have (overline{x}_{m} ightarrow overline{p}) in (E^{n}left(^{*} C^{n} ight)) iff each of the (n) components of (overline{x}_{m}) tends to the corresponding component of (overline{p}), i.e., iff (x_{m k} ightarrow p_{k}, k=1,2, ldots, n,) in (E^{1}(C) .) (See Problem 8 for hints.)

Theorem (PageIndex{3})

Every monotone sequence (left{x_{n} ight} subseteq E^{*}) has a finite or infinite limit, which equals sup_ (_{n} x_{n}) if (left{x_{n} ight} uparrow) and inf (_{n} x_{n}) if (left{x_{n} ight} downarrow .) If (left{x_{n} ight}) is monotone and bounded in (E^{1},) its limit is finite ((b y) 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.

Example (PageIndex{1})

(the number e).

Let (x_{n}=left(1+frac{1}{n} ight)^{n}) in (E^{1} .) By the binomial theorem,

[egin{aligned} x_{n} &= 1+1+frac{n(n-1)}{2 ! n^{2}}+frac{n(n-1)(n-2)}{3 ! n^{3}}+cdots + frac{n(n-1) cdots(n-(n-1))}{n ! n^{n}} [12pt] &= 2+left(1-frac{1}{n} ight) frac{1}{2 !}+left(1-frac{1}{n} ight)left(1-frac{2}{n} ight) frac{1}{3 !}+cdots + left(1-frac{1}{n} ight)left(1-frac{2}{n} ight) cdotsleft(1-frac{n-1}{n} ight) frac{1}{n !} end{aligned}]

If (n) is replaced by (n+1,) all terms in this expansion increase, as does their number. Thus (x_{n}1),

[egin{aligned} 2

Thus (21 .) Hence (2

[2

The following corollaries are left as exercises for the reader.

corollary (PageIndex{1})

Suppose (lim x_{m}=p) and (lim y_{m}=q) exist in (E^{*}).

(a) If (p>q,) then (x_{m}>y_{m}) for all but finitely many (m).

(b) If (x_{m} leq y_{m}) for infinitely many (m,) then (p leq q ;) i.e., (lim x_{m} leq lim y_{m}).

This is known as passage to the limit in inequalities. Caution: The strict inequalities (x_{m}

[x_{m}=frac{1}{m} ext{ and } y_{m}=0.]

Then

yet (lim x_{m}=lim y_{m}=0.)

corollary (PageIndex{2})

Let (x_{m} ightarrow p) in (E^{*},) and let (c in E^{*}) (finite or not). Then the following are true:

(a) If (p>c) (respectively,(pc (x_{m}

(b) If (x_{m} leq c) (respectively, (x_{m} geq c)) for infinitely many (m,) then (p leq c) ((p geq c)).

One can prove this from Corollary (1,) with (y_{m}=c) (or (x_{m}=c)) for all (m).

corollary (PageIndex{3})

(rule of intermediate sequence). If (x_{m} ightarrow p) and (y_{m} ightarrow p) in (E^{*}) and if (x_{m} leq z_{m} leq y_{m}) for all but finitely many (m,) then also (z_{m} ightarrow p).

Theorem (PageIndex{4})

(continuity of the distance function). If

[x_{m} ightarrow p ext{ and } y_{m} ightarrow q ext{ in a metric space } (S, ho),]

then

[ holeft(x_{m}, y_{m} ight) ightarrow ho(p, q) ext{ in } E^{1}.]

Proof

Hint: Show that

[left| holeft(x_{m}, y_{m} ight)- ho(p, q) ight| leq holeft(x_{m}, p ight)+ holeft(q, y_{m} ight) ightarrow 0]

by Theorem 1.