106 CHAPTER 6. MULTI-VARIABLE CALCULUS

Proof: First suppose limn→∞ an = a. Then given ε > 0 there exists nε such that if n> nε ,then

|ank−ak| ≤ |an−a|< ε

which establishes 6.7.8.Now suppose 6.7.8 holds for each k. Then letting ε > 0 be given there exist nk such that

if n > nk,|an

k−ak|< ε/√

p.

Therefore, letting nε > max(n1, · · · ,np) , it follows that for n > nε ,

|an−a|=

(n

∑k=1|an

k−ak|2)1/2

<

(n

∑k=1

ε2

p

)1/2

= ε,

showing that limn→∞ an = a. This proves the theorem.

Example 6.7.4 Let an =(

1n2+1 ,

1n sin(n) , n2+3

3n2+5n

).

It suffices to consider the limits of the components according to the following theorem.Thus the limit is (0,0,1/3) .

Theorem 6.7.5 Suppose {an} and {bn} are sequences and that

limn→∞

an = a and limn→∞

bn = b.

Also suppose x and y are numbers in F. Then

limn→∞

xan + ybn = xa+ yb (6.7.9)

limn→∞

an ·bn = a ·b (6.7.10)

If bn ∈ F, thenanbn→ ab.

Proof: The first of these claims is left for you to do. To do the second, let ε > 0 begiven and choose n1 such that if n≥ n1 then

|an−a|< 1.

Then for such n, the triangle inequality and Cauchy Schwarz inequality imply

|an ·bn−a ·b| ≤ |an ·bn−an ·b|+ |an ·b−a ·b|≤ |an| |bn−b|+ |b| |an−a|≤ (|a|+1) |bn−b|+ |b| |an−a| .

Now let n2 be large enough that for n≥ n2,

|bn−b|< ε

2(|a|+1), and |an−a|< ε

2(|b|+1).