*Then the series
and the improper integral
either both converge or both diverge.*

- , where and is a continuous, non-negative, decreasing function on .
- Moreover:

Thus the given series is convergent.

- Suppose that there exists a convergent series such that, for a given natural number , . Then the series is convergent.
- Suppose that there exists a divergent series such that, for a given natural number , . then the series is divergent.

For , we have . By the integral test (v.s. Thm 3.1 and example 3.2), the series is convergent; thus the given series is convergent.

- If there exists a positive real number L such that , then the two series either are both convergent or are both divergent.
- If and is convergent, then is convergent.
- If and is divergent, then is divergent.

Take . Then:

Thus:

By Thm 2.2 and Thm 9.9, the series is convergent; therefore, by Thm 3.5, the given series is convergent.

- If , the series is convergent.
- If or , the series is divergent.
- If , the test gives no conclusion.

Thus:

As this limit is less than 1, the given series is convergent.

Thus:

By Thm 3.7, the given series diverges.

- If , the series is convergent.
- If or , the series is divergent.
- If , the test gives no conclusion.

We have:

Therefore:

By Thm 3.10, the given series is convergent.

Noah Dana-Picard 2007-12-28