]> Sequences

Math Animated™
Mathematical Introduction for Physics and Engineering
by Samuel Dagan (Copyright © 2007-2020)

Chapter 1: Differentiation; Section 1: Real Numbers; page 3

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Any set of numbers ordered and formed according to a definite rule is called a sequence. The numbers in a sequence are called the terms and each term has an index. The indices consist of natural numbers arranged in increasing order. We will use the notation   a n   for the  nth  term.

The mathematical expression for each term depends on its index  n  and is single valued.

Our concern will be the infinite sequences. It is obvious that an infinite sequence should be countable. We will focus in particular on the converging sequences starting with a simple example:

a n = 5n1 n =5 1 n (

The first few terms, starting from  n = 1  are:   4,    9 2 ,    14 3 ,    19 4 ,   ...

From ( it makes sense to take  5  as the convergence limit of this sequence, which is customary to denote as:

lim n     a n =5 (

A sequence is called convergent, if it has a finite limit. The strict definition of the limit of a sequence is:  A  is the limit of the sequence   a n , if for any positive number  ε  we can find a positive number  N  (depending on  ε) such that  | a n A| < ε  for all integers  n > N . ( | a n A |<ε   is equivalent to:   Aε< a n <A+ε ).

At first this definition looks cumbersome. In order to understand its meaning, we are going to apply it first to the sequence ( and afterwards by graphical means, to a more complicated case.

The definition requires from sequence ( :

| a n A |=| 1 n |= 1 n <εorn> 1 ε (

from ( N should be defined as a number holding the relation:

N 1 ε and thereforen>N 1 ε (

The graphical presentation of N and ε corresponding to the definition is given in: Fig. Convergence of a sequence .

Divergence and Infinity

Any sequence that is not convergent is called divergent, and is not of practical interest for us.

There are however divergent sequences that have a limit which is positive or negative infinity. As an example let's use the sequence:

a n = n 2 (

We write in such a case:

lim n a n = (

The definition for a limit of (positive) infinity is: A sequence   a n   has the limit of infinity if for each positive number  M  one can find a positive number  N  ( depending on  M ) such that for all  n > N ,   a n >M .

One can use the example ( to help understand the definition. Let's take  M = 1000  and therefore our requirement is   a n >1000 . Since   1000 =31.62... , n  should be  32  or biger. One can therefore choose  N = 31.7 , so that for   n32> N ,   a n a 32 =1024>M= 1000 .

There is a similar definition for a limit of   .

Real-number power

On the previous page p.2 Operations an extension of the power to any rational number was obtained. From ( we have:

r p = r n m forr0andp= n m (

On p.1 Irrational Numbers we stated: "It can be shown that between two different real numbers, no matter how close they are, there are at least one rational and one irrational number". This allows to us form a converging sequence of rational numbers to any freely chosen irrational number, and to substitute the irrational number with the limit of the sequence.

The application of this procedure to a sequence of rational powers  p  to any irrational number, allows the extension of  p  to any real number. We therefore have a definition of power

r p forr      and      preal numbers,wherer0 (

As stated earlier on this page, since this definition is obtained by the use of sequences, its terms should be single valued and any number raised to such a power should have a non-negative value.

Practical rules

When a limit of a sequence is calculated, one can obtain one of the undefined expressions given on the previous page p.2 Operations, namely ( and (, which are mentioned here for reference:

{    ;0   ; 0 0    ; 0 0    ; 1    ; 0 } (

In such a case, one has to look for a different approach in order to obtain a defined expression (including infinity). We will see some examples.

The rules presented here also work for infinite limits.

Rules for basic operations

In the case that

A= lim n a n andB= lim n b n (
the following rules apply:
lim n ( a n + b n )=A+B lim n ( a n b n )=AB        lim n ( a n b n )=AB lim n ( a n b n )= A B } (

Rules for operations with power

{ ifR= lim n r n and r n 0 then lim ( r n p )= R p } (

{ ifP= lim n p n andr0 thenlim( r p n )= r P } (

Bounded sequence (the sandwich rule)

One can find in the literature many additional rules. Here is one very useful, which can be proven directly from the definition of convergence:

{ If lim n a n = lim n b n =Cand   there existsNsuch   that   fornN a n c n b n or b n c n a n then lim n c n =C } (

This rule works also for limits of zero and infinity.


All the information necessary for calculating the limits of sequences is given in the previous division "Practical rules", however that is not enough. Ingenuity and experience are also required. In order to solve a problem, it is some times necessary to try different approaches till the solution is achieved. Here are some examples of calculating lim of sequences that can be very useful.

Example 1

In Fig. Convergence of a sequence the following sequence was used:

a n = n 2 +2n3 n 2 3n+7 (

This expression is of type     ( and one cannot use directly the limit of a ratio ( It could be rewritten by multiplying the numerator and the denominator by the common factor   n 2   to obtain:

a n = 1+2 n 1 3 n 2 13 n 1 +7 n 2  

From here after applying the rules ( about limit of a ratio and limit of a sum one obtains: lim n a n =1 . Applying a different factor (e.g. n raised on a power different than −2) won't solve the problem.

Example 2

The following sequence is of type     (, and therefore cannot be solved just by subtraction of limits ( :

a n = 3n(n2) 3n+1 2 n 3 +1 2( n 2 2) (

By bringing the expression to a common denominator one obtains:

a n = 14 n 3 12 n 2 +21n1 2( n 2 2)(3n+1) = 1412 n 1 +21 n 2 n 3 2(12 n 2 )(3+ n 1 )  

where the expression on the right was obtained by a common factor of   n 3 . Finally one obtains:   lim n a n = 7 3 .

Example 3

The following sequence looks may be more complicated, but its solution has some similarities with the two previous:

a n = 2 n+2 6× 2 2n1 3× 2 n1 + 4 n (

There are two  n  dependent factors:   2 n   and   2 2n . In order to be able to convert the term in a more suitable form one should multiply the numerator and the denominator by the factor   2 2n . As a result one obtains:

lim n a n = lim n 4× 2 n 3 3 2 × 2 n +1 =3  

Example 4

This sequence is of type     but needs a different approach than example 2.

a n = n 2 +n n 2 n (

After muliplication and division by:   n 2 +n + n 2 n , one obtains:

lim n a n = lim n 2n n 2 +n + n 2 n = lim n 2 1+ 1 n + 1 1 n =1  

where the common factor   n 1   was applied to the numerator and the denominator.

Example 5

The sequence is of the type   0   ( :

a n = ( 4 n + 3 n ) 1 n (

The ambiguity of   a n   is avoided by requiring   a n >0   for any  n . The term   a n   can be rewritten:

a n = [ 4 n ( 1+ 3 n 4 n ) ] 1 n =4 [ 1+ ( 3 4 ) n ] 1 n  

which leads to  4  multiplied by an expression of type   1 0 , which limit should be one. However there is no rule that allows simultanious calculation of the limit for a number and his power ( In order to calculate the limit, one can use the sandwich rule (

1 1 n < [ 1+ ( 3 4 ) n ] 1 n < 2 1 n  

and one obtains therefore   lim n a n =4 .


Exercise 1. Calculate   lim n c n   if the limit exists and explain why it does not do so for the rest, for the following cases:

c < −1 ; c = −1 ; −1 < c < 0 ; c = 0 ; 0 < c < 1 ; c = 1 ; c > 1

Exercise 2. Calculate

lim n [ 2n( n+3 ) n+1 2n+1 ]  

Exercise 3. Calculate

lim n 10 n+2 +5× 10 2n1 3× 10 n1 + 10 2n  

Exercise 4. Calculate

lim n (   n n 2 3n+2    )  

Exercise 5. Calculate

lim n ( 2 n+3 + 3 n2 ) 1 n  

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