Comparison Tests 

 The Direct Comparison Test

If a series "looks like" a geometric series or a P-series (or some other known series) we can use the test below to determine convergence or divergence.

Theorem:  The Direct Comparison Test  Let            0  <  an   <  bn    for all n (large) If  converges then  also converges. If  diverges then  also diverges.

Example

Determine if 

         

converges.

Solution:  

Let 

       

and 

                      1
        b_n  =               =  e^-n  
                       eⁿ


For  b_n we can use the integral test:


       

We could have also used the geometric series test to show that converges.  
Since converges and

        0  <  a_n  <  b_n  

we can conclude by the comparison test that  converges also.

We use this test when 

                    1
        b_n =              
                    n^p

or other recognizable such as rⁿ.  We often find the b_n by dropping the constants.

Exercises:  Test the following for convergence

Limit Comparison Test

Sometimes, the comparison test does not work since the inequality works the wrong way.  If the functions are similar, then we can use an alternate test.

Limit Comparison Test Suppose that            an > 0,      bn > 0  and                   for some finite positive L. Then both converge or both diverge.

Example:   

Determine if the series 

       

converges.  

Solution

We compare with the harmonic series 

       

which diverges.


We have 

       
Which is a finite positive value.  Thus by the LCT,  diverges.

Exercises:    Determine if the following converge or diverge.

Proof of the Comparison Test

Let a_n   be sequence of positive numbers such that 

        0  <  b_n   <  a_n  

and such that the sequence 

         

then if B_n  represents the nth partial sum of b_n  and A_n is the nth partial sum of a_n then 

        0  <  B_n   <  A_n   <  L 

so that B_n  is a bounded sequence.  B_n is monotonic since the terms are all positive, hence B_n converges.  Now let a_n  be a sequence of positive numbers such that 

        0 < a_n  < b_n  

and such that diverges.  Then if b_n  converges this would contradict the first part of the Comparison test with the roles of a and b switched.  Hence b_n diverges.

Proof of the Limit Comparison Test


Suppose that  diverges and that 

       

then for large n

        a_n   >  b_n (L/2)


but if   diverges so does .  Now by the direct comparison test,  diverges

Notes on the Limit Comparison Test

If   converges and 

       

then a_n  is forced to be very small compared to b_n  for large n and hence  also converges.  

Also if  diverges and 

       

then a_n  is forced to be very large hence  also diverges.  

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