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

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.

- If is convergent, then is convergent.
- If is divergent, then is divergent.

is convergent because:

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

is convergent because:

- the improper integral is convergent (please check it directly!)
- .