Analysis with an introduction to proof 5th by steven lay ch04a

Section 4.2 Limit Theorems... ThenTo simplify our work with convergent sequences, we prove several useful theorems in this section.. The first theorem shows that algebraic operations are

Chapter 4

Sequences

Section 4.2 Limit Theorems

Suppose that (sn) and (tn) are convergent sequences with lim sn = s and lim tn = t Then

To simplify our work with convergent sequences, we prove several useful theorems in this section The first theorem shows that algebraic operations are compatible with taking limits

Theorem 4.2.1

(a) lim (sn + tn) = s + t

(c) lim (sn tn) = s t

(d) lim (sn /tn) = s / t, provided that tn ≠ 0 for all n and t ≠ 0

Proof:

|(sn + tn) – (s + t)| = |(sn – s) + (tn – t)|

(a) For all n ∈ we have

≤ |sn – s| + |tn – t | by the triangle inequality Given any ε > 0,

since sn → s, there exists N1 ∈ such that |sn – s| < ε /2 for all n ≥ N1.

Likewise, there exists N2 ∈ such that |tn – t | < ε /2 for all n ≥ N2.

Now let N = max {N1, N2} Then for all n ≥ N we have

|(sn + tn) – (s + t)| ≤ |sn – s| + |tn – t |

(b) lim (ksn) = ks and lim (k + sn) = k + s, for any k ∈

.

2 2

< + =

Letting M = max {M1, | Thus there exists M1 > 0 such that |  sn | ≤ M1 for all n  t |}, we obtain the inequality

Suppose that (sn) and (tn) are convergent sequences with lim sn = s and lim tn = t Then

Theorem 4.2.1

(c) lim (sn tn) = s t

Proof:

We know from Theorem 4.1.13 that the convergent sequence (sn) is bounded

Now let N = max {N1, N2} Then n ≥ N implies that

Thus, lim sn tn= s t.

Given any ε> 0, there exist natural numbers N1 and N2 such that

| tn – t | <ε /(2M) when n ≥ N1 and | sn – s | <ε /(2M) when n ≥ N2

This time we use the inequality

| sn tn – st | = |(sn tn – sn t) + (sn t – st)|

≤ |(sn tn – sn t)| + |(sn t – st)| = | sn |⋅ | tn – t | + |t |⋅ |sn – s |

.

s t − st ≤ M t − + t M s − s

2 2

Suppose that (sn) and (tn) are convergent sequences with lim sn = s and lim tn = t Then

Theorem 4.2.1

Proof:

Since sn /tn = sn(1/tn), it suffices from part (c) to show that lim (1/tn) = 1/t

That is, given ε> 0, we must make

(d) lim (sn /tn) = s / t, provided that tn ≠ 0 for all n and t ≠ 0

for all n sufficiently large.

To get a lower bound on how small the denominator can be, we note that, since t ≠ 0,

there exists N1 ∈ such that n ≥ N1 implies that |  tn – t | < | t | /2. Thus for n ≥ N1 we have

by Exercise 3.2.6(a)

There also exists N2 ∈ such that n ≥ N2 implies that Let N = max {N1, N2} Then n ≥ N implies that

t t

t − t = t t − < ε

t = − − t t t ≥ t − − t t > t − =

2 1

n

t − < t ε t

.

t − t = t t − = t − t < t − t < ε

Example 4.2.2*

We have sn =

0

5

Now lim (1/n) = 0, so lim (1/n2) = 02 = 0,

lim (– 6 /n2) = (– 6)(0) = 0, and lim [5 – (6 /n2)] = 5.

0

8

Likewise, lim (3/n) = 0, so lim [8 – (3/n)] = 8.

And,

2 2

8

n

−

2 2

n

−

−

2

5 6 /

8 3 /

n n

−

=

−

2 2

8

n

−

Suppose that (sn) and (tn) are convergent sequences with lim sn = s and lim tn = t If sn ≤ tn for all n ∈ , then s ≤ t.

Then ε = (s – t)/2 > 0, and we have 2 ε = s – t and t + ε = s – ε

Thus there exists N1 ∈ such that n ≥ N1 implies that s – ε < sn < s + ε

Theorem 4.2.4

t

ε

s

ε

Similarly, there exists N2 ∈ such that n ≥ N2 implies that t – ε < tn < t + ε

Let N = max {N1, N2} Then for all n ≥ N we have tn < t + ε = s – ε < sn, This contradicts the assumption that sn ≤ tn for all n, and we we conclude that s ≤ t ♦

Corollary 4.2.5

If (tn) converges to t and tn ≥ 0 for all n ∈ , then t ≥ 0

Then since (sn + 1/sn) converges to L, there exists N ∈

such that n ≥ N implies that

A “ratio test” for convergence

Theorem 4.2.7

Suppose that (sn) is a sequence of positive terms and that the sequence of ratios

(sn + 1 / sn) converges to L If L < 1, then lim sn = 0

Let ε = c – L so that ε > 0.

Let k = N + 1 Then for all n ≥ k we have n – 1 ≥ N, so that

It follows that, for all n ≥ k,

Letting M = sk /c k, we obtain 0 < sn < M  cn for all n ≥ k.

Since 0 < c < 1, Exercise 4.1.7(f ) implies that lim cn = 0.

Thus lim sn = 0 by Theorem 4.1.8 ♦

Then there exists a real number c such that 0 ≤ L < c < 1

n n

s

L

s + − < ε

1

n n

s

2

0 < s n < s c n − < s n − c < ×××< s c k n k −

To make 3n/2 > M, we want n > 2M /3.

To handle a sequence such as (sn) = (n) where the terms get larger and larger,

we introduce infinite limits

A sequence (sn) is said to diverge to +∞, and we write lim sn = + ∞ if

for every M ∈ there exists a natural number N such that n ≥ N implies that sn > M.

Show that

Definition 4.2.9

A sequence (sn) is said to diverge to –∞, and we write lim sn = – ∞ if

Example 4.2.11

This time we need a lower bound on the numerator.

We find that 4n2 – 3 ≥ 4n2 – n2 = 3n2, when n ≥ 2.

For an upper bound on the denominator, we have n + 2 ≤ n + n = 2n, when n ≥ 2.

Thus for n ≥ 2 we obtain

So given any M ∈

2

2

n

+

.

+

We conclude with two theorems for infinite limits The proofs are left for the exercises.

Suppose that (sn) and (tn) are sequences such that sn ≤ tn for all n ∈

(a) If lim sn = +∞, then lim tn = + ∞

(b) If lim tn = –∞, then lim sn = –∞.

Let (sn) be a sequence of positive numbers Then

lim sn = + ∞ if and only if lim (1 / sn) = 0

Theorem 4.2.12

Theorem 4.2.13