Comparison Test Part I

I.  Quiz

II.  Homework

III.  The Direct Comparison Test

Theorem:  The Direct Comparison Test

Let 0 < a_n  < b_n  for all n (large)

1)  If sum b_n converges then sum a_n also converges.

2)  If sum a_n diverges then sum b_n also diverges.

We will explain this theorem with a diagram

IV.  Example

Determine if sum 1/exp(n²)  converges.

Solution:  Let a_n = 1/exp(n²) and b_n = 1/eⁿ  = e^-n

For  b_n we can use the integral test:

int from 1 to infinity e^-xdx = -e^-x from 1 to infinity = 1.  We could have also used the geometric series test to show that sum b_n converges.  Since sum b_n converges and

0 < a_n < b_n , we can conclude by the comparison test that sum a_n  converges also.

We use this test when b_n = 1/n^p or other recognizable such as rⁿ.  We often find the b_n by dropping the constants.

Exercises:  Test the following for convergence

A)  sum 1/sqrt(n + 3)

B)  sum 1/(5ⁿ - 6) 

V.  Limit Comparison Test

Suppose that a_n > 0, b_n > 0 and

lim as n -> infinity of a_n /b_n = L (Finite > 0) then both converge or both diverge.

Example:   Determine if the series sum 1/(3n + 5) converges  

We compare with the harmonic series sum 1/n which diverges.

We have lim as n -> infinity of (1/n)/(1/(3n + 5)) = lim as n-> infinity of (3n + 5)/n = lim as n -> infinity of 3/1 = 3.  Thus by the LCT sum 1/(3n + 5) diverges.

Exercises:    Determine if the following converge or diverge.

A)  sum [n² + 3n - 1]/[n3 - 2n + 5]

B)  sum 1/[ln(3n + 2)]