Comparison Test

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.

Example

Determine if 

          \(  \displaystyle\sum_{n=1}^{\infty}\frac{1}{e^{n^2}} \)

converges.

Solution:  

Let 

        \(  \frac{1}{e^{n^2}} \)

and 

                      1
        b_n  =               =  e^-n  
                       eⁿ


For  b_n we can use the integral test:


        \( \int_1^{\infty} e^{-x}dx = e^{-x}|_1^{\infty} = 1 \)

We could have also used the geometric series test to show that \(  \displaystyle\sum_{n=1}^{\infty}b_n \) converges.  
Since \(  \displaystyle\sum_{n=1}^{\infty}b_n \) converges and

        0  <  a_n  <  b_n  

we can conclude by the comparison test that  \(  \displaystyle\sum_{n=1}^{\infty}a_n \) 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

1. \(  \displaystyle\sum_{n=1}^{\infty}\frac{1}{\sqrt{n+3}} \)

2.   \(  \displaystyle\sum_{n=1}^{\infty}\frac{1}{5^n - 6} \)

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.

Example:   

Determine if the series 

        \(  \displaystyle\sum_{n=1}^{\infty}\frac{1}{3n+5} \)

converges.  

Solution

We compare with the harmonic series 

        \(  \displaystyle\sum_{n=1}^{\infty}\frac{1}{n} \)

which diverges.


We have 

        \(  \lim\limits_{n\to \infty} \frac{1}{n} \frac{3n+5}{1} = \lim\limits_{n\to \infty} \frac{3}{1} = 3   \)
Which is a finite positive value.  Thus by the LCT,  \(  \displaystyle\sum_{n=1}^{\infty}\frac{1}{3n+5} \)diverges.

Exercises:    Determine if the following converge or diverge.

1. \(  \displaystyle\sum_{n=1}^{\infty}\frac{2n^2+3n-1}{n^4-2n+5} \)

2. \(  \displaystyle\sum_{n=1}^{\infty}\frac{1}{ln(3n+2)} \)

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 

          \(  \displaystyle\sum_{n=1}^{\infty} a_n = L\)

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 \(  \displaystyle\sum_{n=1}^{\infty} a^n \) 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 \(  \displaystyle\sum_{n=1}^{\infty} b_n \)  diverges and that 

        \(  \lim\limits_{n\to \infty} \frac{a_n}{b_n} = L > 0   \)

then for large n

        a_n   >  b_n (L/2)


but if  \(  \displaystyle\sum_{n=1}^{\infty} b_n \)   diverges so does \( \frac{L}{2} \displaystyle\sum_{n=1}^{\infty} b_n \)  .  Now by the direct comparison test, \(  \displaystyle\sum_{n=1}^{\infty} a_n \)   diverges

Notes on the Limit Comparison Test

If   \(  \displaystyle\sum_{n=1}^{\infty} b_n \)  converges and 

        \(  \lim\limits_{n\to \infty} \frac{a_n}{b_n} = 0 \) 

then a_n  is forced to be very small compared to b_n  for large n and hence  \(  \displaystyle\sum_{n=1}^{\infty} a_n \)  also converges.  

Also if  \(  \displaystyle\sum_{n=1}^{\infty} b_n \)  diverges and 

        \(  \lim\limits_{n\to \infty} \frac{a_n}{b_n} = \infty \) 

then a_n  is forced to be very large hence \(  \displaystyle\sum_{n=1}^{\infty} a_n \)   also diverges.  

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