4.3 Limit of a Sequence: Theorems - KSU Web Home

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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CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES

Example

323

Find

lim

n2 +3n 2n2 +1

.

This problem involves using a standard technique you should remember. We

show all the steps, then we will draw a general conclusion. We begin by factoring

the term of highest degree from both the numerator and denominator.

Now,

n2 + 3n lim 2n2 + 1

=

n2 lim

n2

1

+

3 n

2

+

1 n2

=

lim

1+ 2+

3

n 1 n2

3 lim 1 + n

1 = lim (1) + 3 lim n =1

and

1 lim 2 + n2

1 = lim (2) + lim n2 =2

Since both the limit of the numerator and denominator exist and the limit of the denominator is not 0, we can write

n2 + 3n lim 2n2 + 1

=

lim

1

+

3 n

lim

2

+

1 n2

=1 2

Remark 324 The same technique can be applied to every fraction for which the numerator and denominator are polynomials in n. We see that the limit of such a fraction will be the same as the limit of the quotient of the terms of highest degree. Let us look at some examples:

3n2 + 2n 10

3n2

22

Example 325 lim

= lim = lim n = lim n = 1

2n + 5

2n

33

5n3 2n + 1

5n3

55 1

Example 326 lim 2n4 + 5n2 2 = lim 2n4 = lim 2n = 2 lim n = 0

2n3 n2 + 2n + 1

2n3

Example 327 lim n3 + 10n2 5 = lim n3 = lim 2 = 2

4.3.3 More Theorems on Limits

In example 304 , we used an approximation to simplify the problem a little bit. In this particular example, the approximation was not really necessary, it was

4.3. LIMIT OF A SEQUENCE: THEOREMS

121

more to illustrate a point. Sometimes, if the problem is more complicated, it

may be necessary to use such an approximation in order to be able to ...nd the

condition n has to satisfy. In other words, when we try to satisfy jxn Lj < , we usually simplify jxn Lj to some expression involving n. Let E1 (n) denote this expression. This gives us the inequality E1 (n) < which we have to solve for n. If it is too hard, we then try to ...nd a second expression we will call E2 (n) such that E1 (n) < E2 (n) < . E2 (n) should be such that solving the inequality E2 (n) < is feasible and easier. In order to achieve this, several tricks are used. We recall some useful results, as well as some theorems below.

Theorem 328 (Bernoulli's inequality) If x ber, then (1 + x)n 1 + nx.

1, and n is a natural num-

Theorem 329 (binomial theorem) (a + b)n = an+nan n (n 1) (n 2) an 3b3 + ::: + nabn 1 + bn.

3!

1b+ n (n 2

1) an

2b2+

Corollary 330 (1 + x)n = 1 + nx + n (n 1) x2 + n (n 1) (n 2) x3 + ::: +

2

3!

nxn 1 + xn.

In particular, when x 0, then (1 + x)n is greater than any part of the right hand side. For example, we obtain Bernoulli's inequality: (1 + x)n 1 + nx.

We could also write (1 + x)n n (n 1) x2 or (1 + x)n n (n 1) (n 2) x3.

And so on.

This

is

useful

to

get

2 approximations

on

quantities

3! like

3n.

We

rewrite it as

3n = (1 + 2)n

Theorem 331 (squeeze theorem) If an ! L, cn ! L and an bn cn, then bn ! L Proof. We need to prove that 8 > 09N : n N =) jbn Lj < . Let > 0 be given. Choose N1 such that n N1 =) jan Lj < or < an L < . Similarly, choose N2 such that n N2 =) < cn L < . Let N = max (N1; N2). If n N then

an bn cn () an L bn L cn L () < an L bn L cn L < () < bn L < () jbn Lj <

Theorem 332 If 0 < a < 1 then an ! 0

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CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES

Proof. Let x = 1 1. Then, x > 0 and a = 1 . For n 1,

a

1+x

an

=

1 (1 + x)n

1 by Bernoulli's inequality

1 + nx

1 < nx

To show that an ! 0, we need to show that janj < , for any whenever n N , that is

an < () 1 < nx 1

() n > x

So, given > 0, N; = 1 will work. x

We look at a few more examples, to see how all these results come into play.

2n2 + 1 Example 333 Find lim n2 + 1 using the de...nition Of course, we can ...nd this limit by using the theorems on limits. Here, we do

it using the de...nition, as asked. We think the limit is 2. We want to show that 2n2 + 1

for every > 0, there exists N such that n N =) n2 + 1 2 < . First, we simplify the absolute value.

2n2 + 1 n2 + 1

1 2 = n2 + 1

1 = n2 + 1

1 < n2

1 < if n > 1

n

1

1

So, we see that if < , which happens when n > , then we will have

n

2n2 + 1

1

n2 + 1 2 < . So, N = will work.

1 Example 334 Find lim n2 + 2n 4 We think the limit is 0. We need to prove that for every > 0, there exists N

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