Convergence and Divergence of a Sequence

 

Motivation: 

                       Sometimes, the limit of a sequence can not be found very quickly.  Instead of finding the limit directly, we would like to make sure the limit exists.  To give a flavor of how difficult this kind of problems can be, let’s take a look a famous sequence:

 

                                            where  c_n = ( 1+  )ⁿ 

 

                      The answer for the limit of this sequence will be explored in Calculus. And actually, the limit of the sequence exists.  To describe  the existence of the limit of a sequence instead of finding the limit directly,  a lot of theorems are dedicated to this area.

 

 

Definition ( Convergence and Divergence )

                  If  the limit of a sequence exists,  we say this sequence converges.  If it doe not converge,  we say the sequence diverges.

 

 

 

Example 1:  c_n = (-1)ⁿ .  The limit does not exist.  Thus { c_n  } diverges.

 

 

Example 2:  c_n = 2.  The limit of the sequence is 2.  {c_n } converges.

 

 

 

Theorem ( Cauchy Criterion )

              , c_n  R . 

               The limit of  c_n  exists if and only if  the following statement holds 

 

                        for any   > 0,  there exists an integer  K such that

                                    | c_m – c_n | <    as long as m > K and n> K .

 

    Proof:

 

              (  )

                          Let’s assume the limit of  c_n is L.  In other words,

 

                                 for any  > 0,  there exists an integer K such that

                                        | c_m – L | <    when m > K

                                        | c_n – L | <    when n > K

                              So, 

                                    | c_m – c_n | = | c_m – L + L – c_n |

                                                    | c_m – L | + | L – c_n |

                                                   <  +  =     if m> K, and n> K

 

             (  )

                          The converse part is involved with the completeness of R.  We try not to give a rigorous proof here.  The rigorous proof will be seen in more advanced math.

 

                    

                              for any   > 0,  there exists an integer  K such that

                                      | c_m – c_n | <    as long as m > K and n> K

 

                       And we try to prove the limit of c_n  exists. Let’s consider in this way. Since the statement holds for  m> K and n> K,  we can just choose m to a particular integer larger than K.  Let’s say

 

                                      m=K+1

 

                        Then

                                        | c_n – c_K+1 | <    for n > K

                             -  < c_n – c_K+1 <      and   -  < c_K+1 – c_n <

                             c_K+1 -  < c_n < c_K+1 +

 

                       Let M = max { x  R,  x<c_n for all n except some finitely many terms }

                             ( Strictly speaking, we should use “sup” instead of “max” here.  )

 

                                         M > c_K+1 -  ( because c_n > c_K+1 -  for n > K)

                                              M < c_K+1 +  ( because c_n < c_K+1 +  for n > K )

 

                                 | M – c_K+1 | < 

 

                     Thus,

                                    | M-c_n | < | M – c_K+1 | + | c_K+1 – c_n | <  +  = 2     for n > K

 

                   In other words, we have

                                     for any  > 0,  there exists K such that

 

                                          | M – c_n | < 2   as long as  n> K  .

                   That means  {c_n } converges.

 

 

 

 

 

 

Example:   ,  where c_n = 1 +  +  + … +  .  Determine if it converges.

 

        Sol:

                       For any two integers n > m,  we have

 

                                 c_n – c_m =   +  + … +

                                             <  + … +

                                             =

                                             =  2 ( 2^-(m+1) – 2^-(n+1))

                                             = 2^-m – 2^-n

                                             < 2^-m

 

                       So, for any  > 0,  there exists K  such that

                                       | c_n – c_m | <  | c_n – c_k | < 2^-K <     as long as  n> K and m> K

 

                       Thus,  this sequence converges.