We say

It appears from figure a 7.1 that for every disc centered at the terms of the sequence eventually get into ; i.e., it appears that . Similarly, it appears that .

From figure b, it appears that there are numbers such that , and . You should be able to put your finger on and , and maybe to guess what their exact values are. We will return to these examples later.

Let
. The figure below represents the sequence
. It appears from the figure that there is no number such that
. The following theorem shows that this is the case.
(Note that
.)

Proof: Suppose that and . Then for all
,

In particular,

so

By the triangle inequality,

Combining this result with (7.8), we get

so . This contradiction shows that diverges.

We can also show that constant sequences converge.

Proof: Let
. Let be a disc centered at .
Then

Hence, .

For purposes of calculation it is sometimes useful to rephrase the definition of
convergence. Since the disc is determined by its radius , and for
all
,
, we can reformulate definition
7.6 as