Exercise 5.  Suppose that      has radius of convergence  .  

Show that      has radius of convergence  .  

Solution 5.

See text and/or instructor's solution manual.

Answer.  Show that   .  

Solution.  For   ,      always exists.

For   ,  we have  .  

Use Theorem 4.16 and calculate the    in the Cauchy-Hadamard Formula:    

Details for proofs involving the limit supremum are messy.  So we appeal to the following fact.

Property.  Let    be a sequence of positive real numbers.  The limit supremum ,  

               ,  is the smallest real number  L  with the property that

               there exists a subsequence    and  .

Using this property, we can find a subsequence    such that  

               ,  
               
and it will suffice to use this same subsequence    to compute

                 

Therefore, the radius of convergence of the series    is  .

Remark 1.  The topic of limit supremum, is probably not covered in most calculus courses and this might be your first introduction to it.

Remark 2.  Do not be discouraged there are numerous useful facts that we do not have time to cover in the standard calculus course.

This solution is complements of the authors.

(c) 2008 John H. Mathews, Russell W. Howell