Integral Test and p-Series

The Integral Test

Consider a series S an such that 

        an >    and     an > an+1  

We can plot the points (n,an) on a graph and construct rectangles whose bases are of length 1 and whose heights are of length an.  If we can find a continuous function f(x) such that 

        f(n)  =  an

       
then notice that the area of these rectangles (light blue plus purple) is an upper Reimann sum for the area under the graph of f(x).  Hence 

       


Similarly, if we investigate the lower Reimann sum we see that

       
Since the first term is of a series has no bearing in its convergence or divergence, this proves the following theorem:


           Theorem:  The Integral Test

Let f(x) be a positive continuous function that is eventually 
decreasing and let
f(n) = an
Then
           

converges if and only if

 
           

converges.



 Note the contrapositive:

 Corollary

             

  diverges if and only if 

         
 

  diverges.



Example:

  1. Consider the series:  

           

    We use the Integral Test:

    Let 

            f(x) = 1/x2  

             

    Hence by the Integral Test 

           
     

    converges.

    What does it converge to?  We use a calculator to get 1.635...

  2. Consider the series 

           

    We use the Integral Test.  Let 

             

             

    Hence by the Integral Test 

           

    diverges.

 


P-Series Test

A special case of the integral test is when  

                     1
        an  =            
                    np 

for some p.  The theorem below discusses this.

        Theorem:  P-Series Test
Consider the series 

                  

  1. If p > 1 then the series converges

  2. If 0 < p < 1 then the series diverges

 

Proof:

We use the integral test with the function 

                        1
        f(x)  =            
                       xp 

For p not equal to 1

       

Note that this limit converges if

        -p + 1 < 0 

or 

        p > 1

The limit diverges for p < 1

For p = 1 we have the harmonic series

       

and the integral test gives:

       

Another proof that the harmonic series diverges.

 



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