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Limits of Sequences of Real Numbers

Sequences of Real NumbersLimits through Rigorous

DefinitionsThe Squeeze Theorem

Using the Squeeze TheoremMonotonous Sequences

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Sequences of Numbers

1 2 3A ,x ,x , is a rule that assigns,

to each natural number , t

sequenc

he numb

e

.

er n

x

n x

1 1 1

1, , , ,2 4 8

1,1.4,1.41,1.414,1.4142,

Definition

Examples 1

2

1, 3,5, 7,9,3

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Limits of Sequences

1 2 3

A finite number is the of the sequence

,x ,x , if the numbers get arbitrarily close

to the number as the ind

li

ex gro

t

w

mi

.

sn

L

x x

L n

1 1 1The sequence 1, , , , converges

2 4 8

and its limit is 0.

Definition

Examples 1

If a sequence has a finite limit, then we say that the sequence is convergent or that it converges. Otherwise it diverges and is divergent.

0

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1 1 1The sequence 1, , , , converges

2 4 8

and its limit is 0.

0

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Limits of Sequences

The sequence 1,1.4,1.41,1.414,1.4142, converges

and its limit is 2.

2

3

Notation lim nnx L

The sequence (1,-2,3,-4,…) diverges.

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Computing Limits of Sequences (1)

The limit of a sequence can be often computed by inserting

in the formula defining the general term . If this expression can be

evaluated and the result is finite, then this finite value is

n

n

x n

x

the limit of

the sequence. This usually requires a rewriting of the expression . nx

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Computing Limits of Sequences (1)

1

1

1 1 1 1The limit of the sequence 1, , , , is 0 because

2 4 8 2

1inserting to the formula one gets 0.

2

n

n nn x

2 2 2

2 2

2

111 1

The limit of the sequence is 1 because rewriting 11 1 1

and inserting one gets 1.

n n nn n

nn

Examples

1

2

1n2

0

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Computing Limits of Sequences

The limit of the sequence 1 is 0 because of the rewriting n n

1 11

1

n n n nn n

n n

Insert to get the limit 0.n

Examples continued

3

1 1 .

1 1

n n

n n n n

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Formal Definition of Limits of Sequences

1 2 3

A finite number is the of the sequence

, , , if

0 : such that

lim t

.

i

n

L

x x x

n n n L x

Definition

Example1

lim 0 since if 0 is given, thenn n

1 1 10 if .n n

n n

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Limit of Sums

Assume that the limits lim and lim

are finite. Then lim .

n nn n

n nn

x x y y

x y x y

Let 0 be given.

To that end observe that also 0.2

Theorem

Proof

We have to find a number with the property

.n n

n

n n x y x y

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Limit of Sums

Proof1 2

1 2

Hence there are numbers and such that

and .2 2n n

n n

n n x x n n y y

1 2Let now =max , . We have

.2 2n n n n

n n n

n n x y x y x x y y

By the Triangle Inequality

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Limits of Products

2

1Let 1 and . Then lim 0 and

the limit lim does not exist. However, lim 0.

n

n n nn

n n nn n

x n y yn

x x y

The same argument as for sums can be used to prove the following result.

Assume that the limits lim and lim

are finite. Then lim .

n nn n

n nn

x x y y

x y x y

Theorem

Remark

Examples

Observe that the limits lim and lim may exist

and be finite even if the limits lim and lim do not exist.

n n n nn n

n nn n

x y x y

x y

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Squeeze Theorem for Sequences

Then the limit lim exists and

lim lim lim .

nn

n n nn n n

y

y x z

Assume that : and that

lim lim .n n n

n nn n

n x y z

x z a

Theorem

Let max , . Then

max , .

y x z

y n n n

n n n

n n a y a x a z

Let 0. Since lim lim , such

that and .

n n x zn n

x n z n

x z a n n

n n x a n n z a

Proof

This follows since .n n nx y z n

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Using the Squeeze Theorem

Next observe that

1 2 3 1! 1 2 3 1

n

n nn n n

n n n n n n n n n n n

!Compute lim .nn

nn

Example

! 1Hence 0 .n

nn n

!Observe that 0< for all 0.n

nn

n

Solution This is difficult to compute using the standard methods because n! is defined only if n is a natural number.

So the values of the sequence in question are not given by an elementary function to which we could apply tricks like L’Hospital’s Rule.

1 !Since lim 0, also lim 0 by the Squeeze Theorem.nn n

nn n

Here each term k/n < 1.

1

.n

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Using the Squeeze Theorem

sin( )Does the sequence converge?

cos( )

If it does, find its limit.

n

n nProblem

Solution

1 1Since lim lim 0 we conclude that the sequence

-1 -1

sin( ) sin( ) converges and that lim 0.

cos( ) cos( )

n n

n

n n

n n

n n n n

1 sin( ) 1

Hence .1 cos( ) 1

n

n n n n

We have 1 sin( ) 1 and 1 cos( ) 1 for all 2,3,4, . n n n

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Monotonous SequencesDefinition

The sequence (a1,a2,a3,…) is decreasing if an+1 ≤ an for all n.

A sequence (a1,a2,a3,…) is increasing if an ≤ an+1 for all n.

The sequence (a1,a2,a3,…) is monotonous if it is either increasing or decreasing.

Theorem

The sequence (a1,a2,a3,…) is bounded if there are numbers M and m such that m ≤ an ≤ M for all n.

A bounded monotonous sequence always has a finite limit.

Observe that it suffices to show that the theorem for increasing sequences (an) since if (an) is decreasing, then consider the increasing sequence (-an).

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Monotonous SequencesTheorem A bounded monotonous sequence always has a finite limit.

Proof Let (a1,a2,a3,…) be an increasing bounded sequence.

Then the set {a1,a2,a3,…} is bounded from the above.

By the fact that the set of real numbers is complete, s=sup {a1,a2,a3,…} is finite.

lim .nna s

Claim

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Monotonous SequencesTheorem A bounded monotonous sequence always has a finite limit.

Proof Let (a1,a2,a3,…) be an increasing bounded sequence.

Let s=sup {a1,a2,a3,…}.

lim .nna s

Claim

Proof of the Claim Let 0.

We have to find a number with the property that .nn n n a s

.Since sup , there is an element such that n n n ss a a s a

Since is increasing .n n na n n s a a s

Hence .nn n a s This means that lim .nna s