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


          0  <  a <  b  

for all n (large)

  1. If  converges then  also converges.

  2. If  diverges then  also diverges.




Determine if 







        bn  =               =  e-n  

For  bn we can use the integral test:


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

        0  <  an  <  bn  

we can conclude by the comparison test that  converges also.


We use this test when 

        bn =              

or other recognizable such as rn.  We often find the bn 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 


for some finite positive L. Then both converge or both diverge.


Determine if the series 





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 an   be sequence of positive numbers such that 

        0  <  b <  an  

and such that the sequence 


then if Bn  represents the nth partial sum of bn  and An is the nth partial sum of an then 

        0  <  B <  An   <  L 

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

        0 < an  < bn  

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

Proof of the Limit Comparison Test

Suppose that  diverges and that 


then for large n

        an   >  bn (L/2)

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

Notes on the Limit Comparison Test

If   converges and 


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

Also if  diverges and 


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


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