Convergence theorems.

First we give a very important improper integral, which will be used together with the tests for convergence.

Proposition 8.9.9   The improper integral

is convergent for and is divergent for .

Proof. For , we have:

Then:

This limit is infinite for , and is equal to 1 for .

For , we have:

This integral has infinite limit for in a neighborhood of , i.e. the given improper integral diverges.

Proposition 8.9.10 (Direct Comparison)   Let and be two functions, continuous on the interval . Suppose that for every , they verify . Then:
1. If is convergent, then is convergent.
2. If is divergent, then is divergent.

Example 8.9.11   Let , for . The improper integral

is convergent because:
• For any , we have .
• The improper integral is convergent (by Prop. 9.9).

Proposition 8.9.12 (Limit Comparison)   Let and be two functions, continuous and positive on the interval . Suppose that there exists a real positive number such that

Then the improper integrals and are either both convergent or both divergent.

Example 8.9.13   The improper integral

is convergent because:
• the improper integral is convergent (please check it directly!)
• .

Noah Dana-Picard 2007-12-28