*Proof*.
We will only prove 1.
Assume that we have

. Let

, and notice
that

(since

, so

, so

,
and also

).

Since
, there is
an such that for all we have

so

Then we have

Here the common ratio for the second one is

, hence
thus the right-hand series converges, so the left-hand
series converges.

**Example 6.4.7**
Consider

.
The ratio of successive terms is

Thus this series converges

*absolutely*.
Note, the minus sign is missing above since in the ratio test
we take the limit of the absolute values.

**Example 6.4.9**
Let's apply the ratio test to

.
We have

This tells us nothing.
If this happens... do something else! E.g., in this case, use the
integral test.