Pc Intro To Sequences

Limits of Sequences of Real Numbers Sequences of Real Numbers Limits through Rigorous Definitions The Squeeze Theorem Using the Squeeze Theorem Monotonous Sequences

Sequences of Numbers Definition

A sequence

 x1,x2 ,x3 , K 

is a rule that assigns,

to each natural number n, the number xn. Examples

1



1 1 1  1, , , ,K   2 4 8  

2

 1,1.4,1.41,1.414,1.4142,K 

3

 1, 3,5, 7,9,K 

Limits of Sequences Definition

A finite number L is the limit of the sequence

 x1,x2 ,x3 , K 

if the numbers xn get arbitrarily close

to the number L as the index n grows.

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.  1 1 1  The sequence 1, , , , K 1 Examples  2 4 8  converges  

and its limit is 0. 2

The sequence  1,1.4,1.41,1.414,1.4142, K and its limit is

3

2.

The sequence (1,-2,3,-4,…) diverges.

Notation

lim xn  L n 



converges

Computing Limits of The limit of a sequence  x  (1) can be often computed by inserting n   Sequences in the formula defining the general term x . If this expression can be n

n

evaluated and the result is finite, then this finite value is the limit of the sequence. This usually requires a rewriting of the expression xn .  1 1 1   1  The limit of the sequence 1 Examples  1, 2 , 4 , 8 ,K    2n 1  is 0 because     1 inserting n   to the formula xn  n 1 one gets 0. 2 1 1  2 2 2  n  1 n 1 n 2 The limit of the sequence  2   is 1 because rewriting 2 1 n  1 n  1   1 2 n and inserting n   one gets 1.

Computing Limits of Sequences Examples continued 3

The limit of the sequence

n 1

 n 





n  1  n is 0 because of the rewriting

n 1 n



n 1 n

n 1 n

 n  1  n n 1 n



1 n 1 n

Insert n   to get the limit 0.

.



Computing Limits by Maple Maple commands Limit and limit Calling Sequence

Limit(f,x=a,dir) and limit(f,x=a,dir)

This command computes the limit of the expression f as the variable x approaches the value a. The optional argument dir can be used to define the direction from which the variable x approaches the value a. When computing limits of sequences, f is the general term of the sequence and the variable x takes only positive integer values and approaches the infinity.

Formal Definition of Limits of Sequences Definition

A finite number L is the limit of the sequence

 x1,x2 ,x3 , K 

if

  0 : n such that n  n  L  xn   .

Example

1  0 since if   0 is given, then n  n

lim

1 1 1 0    if n   n . n n 

Limit of Sums Theorem

Assume that the limits lim xn  x and lim y n  y  n 

are finite. Proof

Then

n 

lim  xn  y n   x  y  .

n 

Let   0 be given. We have to find a number n with the property n  n  xn  y n  x  y    .

To that end observe that also

  0. 2

Hence there are numbers n1 and n2 such that n  n1  xn  x 

  and n  n2  y n  y   . 2 2

Let now n =max  n1, n2  . We have

n  n  xn  y n  x  y   xn  x  y n  y  

By the Triangle Inequality

    . 2 2

Limits of Products The same argument as for sums can be used to prove the following result. Theorem

Assume that the limits lim xn  x and lim y n  y  n 

are finite. Remark

Then

n 

lim xn y n  x y  .

n 

Observe that the limits lim xn y n and lim  xn  y n  may exist n 

n 

and be finite even if the limits lim xn and lim y n do not exist. n 

n 

1 . Then lim y n  0 and 2 n  n the limit lim xn does not exist. However, lim xn y n  0.

Examples Let xn   1 n and y n  n

n 

n 

Squeeze Theorem for Sequences Theorem Assume that n : x  y  z and that n

n

n

lim xn  lim zn  a.

n 

n 

Then the limit lim y n exists and n 

lim y n  lim xn  lim zn .

n 

Proof

n 

n 

Let   0. Since lim xn  lim zn  a, nx  nz such n 

n 

that n  nx  xn  a   and n  nz  zn  a   .

Let ny  max  nx , nz  . Then

n  ny  a  y n  max  a  xn , a  zn    .

This follows since xn  y n  zn n.

Using the Squeeze Theorem Example Solution

n! . n  nn

Compute lim

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. n! Here each term k/n < 1. Observe that 0< n for all n  0. n Next observe that

n ! 12 3L  n  1 n 1 2 3 n  1 n 1     L   . n n n n n L n n n n n n n n

Hence 0 

n! 1  . n n n

1 n!  0, also lim n  0 by the Squeeze Theorem. n  n n  n

Since lim

Using the Squeeze Theorem 

sin(n )   converge?  n  cos(n )  If it does, find its limit. Does the sequence 

Problem Solution

We have

 1  sin( n )  1

Hence

1 sin(n ) 1    . n  1 n  cos(n ) n  1

and

 1  cos(n )  1 for all n  2,3,4,K .

1 1    lim    0 we conclude that the sequence  n  n - 1 n   n - 1



Since lim

sin(n )  sin(n ) converges and that lim  0.   n  n  cos(n )  n  cos(n ) 

Monotonous Sequences Definition

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

The sequence (a1,a2,a3,…) is decreasing if an+1 ≤ an for all n. The sequence (a1,a2,a3,…) is monotonous if it is either increasing or decreasing. The sequence (a1,a2,a3,…) is bounded if there are numbers M and m such that m ≤ an ≤ M for all n. Theorem

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).

Monotonous Sequences Theorem Proof

A bounded monotonous sequence always has a finite limit. 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. Claim

lim an  s. n 

Monotonous Sequences Theorem Proof

A bounded monotonous sequence always has a finite limit. Let (a1,a2,a3,…) be an increasing bounded sequence.

Let s=sup {a1,a2,a3,…}. Claim

lim an  s. n 

Proof of the Claim

Let   0.

We have to find a number n with the property that n  n  an  s   . Since s  sup  an  , there is an element an such that s    an  s. Since  an  is increasing n  n  s    an  an  s. Hence n  n  an  s   . This means that lim an  s. n 