4.3 Limit of a Sequence: Theorems

4.3. LIMIT OF A SEQUENCE: THEOREMS

115

4.3 Limit of a Sequence: Theorems

These theorems fall in two categories. The ...rst category deals with ways to combine sequences. Like numbers, sequences can be added, multiplied, divided, ... Theorems from this category deal with the ways sequences can be combined and how the limit of the result can be obtained. If a sequence can be written as the combination of several "simpler" sequences, the idea is that it should be easier to ...nd the limit of the "simpler" sequences. These theorems allow us to write a limit in terms of easier limits. however, we still have limits to evaluate. The second category of theorems deal with speci...c sequences and techniques applied to them. Usually, computing the limit of a sequence involves using theorems from both categories.

4.3.1 Limit Properties

We begin with a few technical theorems. They do not play an important role in computing limits, but they play a role in proving certain results about limits.

Theorem 310 Let x be a number such that 8 > 0, jxj < , then x = 0. Proof. See problems at the end of the section.

Theorem 311 If a sequence converges, then its limit is unique.

Proof. We assume that an ! L1 and an ! L2 and show that L1 = L2. Given

> 0 choose N1 such that n

N1 =) jan

L1j <

. 2

Similarly, choose N2

such that n

N2 =) jan

L2j

<

. 2

Let

N

=

max (N1; N2).

If

n

N , then

jL1 L2j = jL1 an + an L2j jan L1j + jan L2j

0 such that janj M for all n. Proof. Choose N such that n N =) jan Lj < 1. By the triangle inequality, we have

janj jLj jan Lj 09N : m; n jan amj <

N =)

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CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES

Proof. Given > 0, we can choose N such that n; m

and jam

Lj < . Now, 2

N =) jan Lj < 2

jan amj = jan L + L amj = j(an L) (am L)j j(an L)j + jam Lj by the triangle inequality

0, there exists N such that k N =) jbk Lj < . Let > 0 be given. Choose N such that nk N =) jank Lj < . Now, if k N , then nk N therefore

jbk Lj = jank Lj <

This theorem is often used to show that a given sequence diverges. To do so, it is enough to ...nd two subsequences which do not converge to the same limit. Alternatively, once can ...nd a subsequence which diverges.

Example 316 Study the convergence of cos n The subsequence cos 2n converges to 1, while the subsequence cos (2n + 1) converges to 1. Thus, cos 2n must diverge.

In the next two sections, we look at theorems which give us more tools to compute limits.

4.3. LIMIT OF A SEQUENCE: THEOREMS

117

4.3.2 Limit Laws

The theorems below are useful when ...nding the limit of a sequence. Finding the

limit using the de...nition is a long process which we will try to avoid whenever

possible. Since all limits are taken as n ! 1, in the theorems below, we will

write

lim an

for

lim

n!1

an

.

Theorem 317 Let (an) and (bn) be two sequences such that an ! a and bn ! b with a and b real numbers. Then, the following results hold:

1. lim (an bn) = (lim an) (lim bn) = a b.

2. lim (anbn) = (lim an) (lim bn) = ab.

3. if lim bn = b 6= 0 then lim

an bn

=

lim an

=

a .

lim bn b

4. lim janj = jlim anj = jaj.

5. if an 0 then lim an 0.

6. if an bn then lim an lim bn.

7. if lim an = a

0

then

lim pan

=

p lim an

=

p a.

Proof. We prove some of these items. The remaining ones will be assigned as problems at the end of the section.

1. We prove lim (an + bn) = (lim an) + (lim bn). The proof of lim (an bn) = (lim an) (lim bn) is left as an exercise. We need to prove that 8 > 0, 9N : n N =) jan + bn (a + b)j < . Let > 0 be given, choose N1

such that n

N1 =) jan

aj <

. 2

Choose N2 such that n

N2 =)

jbn

bj

<

. 2

Let

N

=

max (N1; N2).

If

n

N , then

jan + bn

(a + b)j = j(an a) + (bn b)j jan aj + jbn bj by the triangle inequality

0, 9N : n N =) janbn abj < . Since an converges, it is bounded, let M be the bound i.e. janj < M . Choose

N1 such that n

N1 =) jan

aj <

. 2 (jbj + 1)

Choose N2 such that

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CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES

n

N2 =) jbn

bj

<

2 (M

. + 1)

Let

N

=

max (N1; N2).

If

n

N then

janbn

abj = janbn anb + anb abj = jan (bn b) + b (an a)j janj jbn bj + jbj jan aj

< M 2 (M + 1) + jbj 2 (jbj + 1) 0, 9N : n jjanj jajj < . Let > 0 be given, choose N such that n jan aj < (since an ! a) . If n N , then we have:

N =) N =)

jjanj jajj < jan aj by the triangle inequality <

5. We prove it by contradiction. Assume that an ! a < 0. Choose N such

1

that n

N ) jan

aj <

a. Then, 2

1 an a < 2 a

1

)

an

<

a 2

) an < 0

which is a contradiction.

6. We apply the results found in parts 1 and 5 to the sequence an bn.

7. See problems

Remark 318 Parts 1, 2 and 3 of the above theorem hold even when a and b are extended real numbers as long as the right hand side in each part is de...ned. You will recall the following rules when working with extended real numbers:

1. 1 + 1 = 1 1 = ( 1) ( 1) = 1 2. 1 1 = ( 1) 1 = 1 ( 1) = 1 3. If x is any real number, then

(a) 1 + x = x + 1 = 1

4.3. LIMIT OF A SEQUENCE: THEOREMS

119

(b) 1 + x = x 1 = 1

(c) x = x = 0 11

(d) x = 0

1 if x > 0 1 if x < 0

(e) 1 x = x 1 =

1 if x > 0 1 if x < 0

(f ) ( 1) x = x ( 1) =

1 if x > 0 1 if x < 0

4. However, the following are still indeterminate forms. Their behavior is unpredictable. Finding what they are equal to requires more advanced techniques such as l'H?pital's rule.

(a) 1 + 1 and 1 1

(b) 0 1 and 1 0 (c) 1 and 0

10

Remark 319 When using theorems from this category, it is important to remember previous results since these theorems allow us to write a limit in terms of other limits, we hopefully know. The more limits we know, the better o? we are.

Example 320 If

section

that

lim

1 n

c

6=

0,

...nd

lim

c n

.

= 0. Therefore

We

know

from

an

example

in

the

previous

c lim

= c lim 1

n

n

=c 0

=0

Example

321

Find

lim

1 n2

.

In the previous section, we computed this limit

using the de...nition. We can also do it as follows.

1

1

lim n2 = lim n

1 = lim

n =0 0

=0

1 n

1 lim

n

Remark 322 From the above example, we can see that if p is a natural number,

lim

1 np

=

0.

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