Kenneth Kuttler

4.4. THE LIMIT OF A SEQUENCE 59

F× ·· ·×F where F consists of real or complex numbers and, although other definitionsare used,

∥a∥ ≡max{|ai| : i≤ p}for a = (a1, · · · ,ap) ∈ Fp. In this book, it is usually the case that p = 1, but there is nodifficulty in considering a more general case.

Proposition 4.4.2 The usual properties of absolute value hold for ∥·∥ with addition ofthe vectors, and multiplication by a scalar α, as presented in elementary calculus

a+b≡ (a1 +b1, · · · ,ap +bp) , αa≡ (αa1, · · · ,αap)

Proof: From the triangle inequality for complex numbers,

∥a+b∥ ≡max{|ai +bi| , i≤ p} ≤max{|ai| , i≤ p}+max{|bi| , i≤ p}= ∥a∥+∥b∥ .(4.2)

Also, for α ∈ F, ∥αa∥ ≡ max{|αai| : i≤ p} = |α|max{|ai| : i≤ p} = |α|∥a∥. By defi-nition, ∥a∥ ≥ 0 and is 0 if and only if ai = 0 for each i if and only if a = 0 ≡ (0, · · · ,0) .Also ∥a∥ = ∥a−b+b∥ ≤ ∥a−b∥+ ∥b∥ so ∥a∥−∥b∥ ≤ ∥a−b∥ . Similarly ∥b∥−∥a∥ ≤∥b−a∥= ∥a−b∥ and so |∥a∥−∥b∥| ≤ ∥a−b∥.

In words the definition says that given any measure of closeness ε, the terms of thesequence are eventually this close to a. Here, the word “eventually” refers to n being suffi-ciently large. The above definition is always the definition of what is meant by the limit ofa sequence.

Proposition 4.4.3 Let an =(an

1, · · · ,anp). Then an → a if and only if for each i ≤ p,

ani → ai.

Proof:⇒ is obvious because |ani −ai| ≤ ∥an−a∥.

⇐ There exists ni such that∣∣ak

i −ai∣∣< ε whenever k > ni. Let

N ≥max{ni : i≤ p}

Then for n≥ N,∥an−a∥ ≡max{|ani −ai| , i≤ p}< ε .

Theorem 4.4.4 If limn→∞ an = a and limn→∞ an = â then â = a.

Proof: Suppose â ̸= a. Then let 0 < ε < ∥â−a∥/2 in the definition of the limit. Itfollows there exists nε such that if n≥ nε , then ∥an−a∥< ε and ∥an− â∥< ε. Therefore,for such n, ∥â−a∥ ≤ ∥â−an∥+∥an−a∥< ε + ε < ∥â−a∥/2+∥â−a∥/2 = ∥â−a∥, acontradiction.

Example 4.4.5 Let an =(

1n2+1 ,

)∈ F2.

Then it seems clear that limn→∞ an = (0,0) . In fact, this is true from the definition.Let ε > 0 be given. Let nε ≥ max

(√ε−1, 1

). Then if n > nε ≥

√ε−1, it follows that

n2 +1 > ε−1 and so 0 < 1n2+1 = an < ε and also n≥ 1/ε so 1/n < ε . Thus ∥an− (0,0)∥ ≡

max(∣∣∣ 1

n2+1 −0∣∣∣ , ∣∣ i

n −0∣∣)< ε whenever n is this large.

Note the definition was of no use in finding a candidate for the limit. This had to beproduced based on other considerations. The definition is for verifying beyond any doubtthat something is the limit. It is also what must be referred to in establishing theoremswhich are good for finding limits.

4.4. THE LIMIT OF A SEQUENCE 59F x --- x F where F consists of real or complex numbers and, although other definitionsare used,\la|| = max {|aj| i < p}for a= (a,---,ap) € F?. In this book, it is usually the case that p = 1, but there is nodifficulty in considering a more general case.Proposition 4.4.2 The usual properties of absolute value hold for ||-|| with addition ofthe vectors, and multiplication by a scalar Q, as presented in elementary calculusa+b=(a,+bh1,::-,dpt+bp), ha = (Qa,--- , Ady)Proof: From the triangle inequality for complex numbers,\|a + b|| = max {|a;+ bi] 7 < p} < max {|ai| 7 < p} + max {|bi i < p} = lla] + |||).(4.2)Also, for a € F, ||@a|| = max {|aa;| :i < p} = |a|max {|a;|:i< p} = |a||la||. By defi-nition, ||a|| > 0 and is 0 if and only if a; = 0 for each i if and only if a= 0 = (0,--- ,0).Also ||al| = j]a—b +d] < |la—d|| + |]5I| so lal] — |||] < |la— 4]. Similarly {||| — lal] <\|b —al| = ||a— || and so ||la|| — ||| < la—4|. 0In words the definition says that given any measure of closeness €, the terms of thesequence are eventually this close to a. Here, the word “eventually” refers to n being suffi-ciently large. The above definition is always the definition of what is meant by the limit ofa sequence.Proposition 4.4.3 Let a, = (af, ve a) . Then ay, — a if and only if for each i < p,a > dj.Proof: = is obvious because |a? — aj| < ||dn — all.< There exists n; such that |at — a;| < € whenever k > nj;. LetN > max {n;:i< p}Then for n > N, ||a, —a|| = max {|a? —aj|,i< p}<e.Theorem 4.4.4 Tf limy soo dy = a and liMy4.0 dy = G then 4G = a.Proof: Suppose d 4 a. Then let 0 < € < ||G@—al| /2 in the definition of the limit. Itfollows there exists ng such that if n > ne, then ||a, —al| < € and ||a, — a|| < €. Therefore,for such n, ||4—al| < |]J4@—ap|| + lan — al] < e+ < ||@—al| /2+ ||4—al| /2 = ||@—all, acontradiction. JjExample 4.4.5 Let ay = ( ! ‘) eF’.m+l?nThen it seems clear that limy_;.0d, = (0,0). In fact, this is true from the definition.Let € > 0 be given. Let ng > max (v el, 1) . Then if n > ng > Veo, it follows thatn?+1>e7!andso0< az =a, < € and alson > 1/€ so 1/n < €. Thus ||a, — (0,0) || =1—0n+Note the definition was of no use in finding a candidate for the limit. This had to beproduced based on other considerations. The definition is for verifying beyond any doubtthat something is the limit. It is also what must be referred to in establishing theoremswhich are good for finding limits.max (,|E- 0) < € whenever n is this large.