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

converges if and only if



 Note the contrapositive:



  diverges if and only if 




  1. Consider the series:  


    We use the Integral Test:


            f(x) = 1/x2  


    Hence by the Integral Test 



    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 




P-Series Test

A special case of the integral test is when  

        an  =            

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



We use the integral test with the function 

        f(x)  =            

For p not equal to 1


Note that this limit converges if

        -p + 1 < 0 


        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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