Absolute Convergence and Ratio Test for Series

 

Motivation: 

                      Previously, the monotone convergence theorem has been introduced. The result of that theorem can be applied on series.  For example,  for a series

     

                                       S_n = ,   where c_i > 0

 

                     If  we have  | S_n | < M,  then we know the limit of  S_n exists from the result of monotone convergence theorem.  Based on that simple result, some tools can be developed to check if a series is convergent or not.

 

 

 

Definition ( Absolute Convergence )

                  A series is known as

                                                       ,  n  N .

                 If   ,  then we say that the series converges absolutely.

 

 

 

 

 

Theorem :  

        , n  N .  If the series converges absolutely, then the limit of S_n exists.

 

        Proof:

                        Let   .   Since the limit of  A_n exists, we have

                            for any  > 0,  there exists K such that

                                   | A_n – A_m | <   if  n> m>  K    ( Cauchy criterion )

 

                       And  | A_n – A_m | = | c_m+1 | + | c_m+2 | + … + | c_n |

                                                   | c_m+1 + c_m+2 + … + c_n |  = | S_n – S_m |  

 

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

                              | S_n – S_m | <    if   n>m> K

 

                     In other words, the limit of S_n exists.

 

 

 

Theorem :   , n  N .  If  { S_n } converges,  then   .

   Proof:

                   { S_n } converges.  Then for any  > 0,  there exists K such that

                                 | S_n – S_m | <   if  n > m > K   ( Cauchy criterion )

 

                  In particular,  we can have  n > (n-1) > K such that

                              | c_n | = | S_n – S_n-1 |  <    for  n > K

 

                  That means   .

 

Note:  The converse of this theorem is not always true.   does not guarantee  that  the series will converge.

 

 

 

 

 

 

 

Theorem ( Ratio Test for Series )

        ,  n  N .   Let   .   Then

(1)   If  r< 1,  then  {S_n} converges.

(2)   If r > 1,  then { S_n | diverges.

(3)   When r=1, this test does not give any information.

 

      Proof:

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

                                          <    if  n > K .

                        |c_n | (r-  ) < | c_n+1 | < | c_n | (  + r )     for  n > K .

                     

                          | c_K | (r-  )^m  < | c_K+m | < | c_K | (  +r )^m 

 

                (1)     Hence,  if r < 1, we can choose  such that  (r+  ) < 1.  Then

 

                                   =

                                               <  <  even when m goes to  .

 

                  So,       { S_n } converges.

 

               (2)   if    > 1,   then  we can choose  such that  (r-  ) > 1 .  Then

                           | c_K | (r-  )^m  < | c_K+m | 

                           | c_n | > | c_K |   for every n > K . 

                 

                     Since | c_K | is not equal to 0, there is no hope that  { S_n } will converge.

                    { S_n } diverges.

 

 

 

 

 

 

Example:   ,    .  Check if {S_n} converges.

        Sol:

                      

                    So,    .    Thus,  { S_n } converges.