# Divergent Sequences

Example 1.
Prove the sequence
$\{a_n\}$
​ given by
$a_n \triangleq (-1)^n$
​ for all
$n\in\mathbb{N}$
​ diverges.
Comment
Hint
Visual
Formal Write-Up
Follow-Up
Numerous notions of "diverge" in among academic texts. In our setting, divergence is meant in the sense of the sequence failing to converge.
Can you pick a value for
$\varepsilon$
​ that would prove, by way of contradiction, the limit definition does not hold?
The sequence oscillates between
$\mathsf{-1}$
and
$\mathsf{1}$
. Letting ​
$\mathsf{\epsilon=1/2}$
$\{a_n\}$
​ converges. This implies there is
$\alpha\in\mathbb{R}$
​ and
$N\in\mathbb{N}$
​ such that
$|a_n - \alpha|\leq 1/2$
​ for all
$n\geq N$
​. In particular,
|
$\dfrac{1}{2} \geq |a_{2N} - \alpha| = |1-\alpha| \geq 1-\alpha \quad \implies\quad \alpha \geq \dfrac{1}{2}$
| | | ---------------------------------------------------------------------------------------------------------------- | :-: |
​and
|
$\dfrac{1}{2} \geq |a_{2N+1} - \alpha| = |(-1)-\alpha| \geq -1-\alpha \quad \implies \quad \alpha \leq -\dfrac{1}{2}.$
| | | ------------------------------------------------------------------------------------------------------------------------ | :-: |
Combining these inequalities reveals
 ​$\dfrac{1}{2} \leq \alpha \leq -\dfrac{1}{2}$​ ​
a contradiction. Thus, the initial assumption
$\{a_n\}$
​ converges was false, and we are done.
 ​$\blacksquare$​ ​
To tackle this problem, we first label
$\{a_n\}$
, which conveniently only has two values on the number line. If something converges, then iterates eventually get and stay close to some limit
$\alpha$
. However, here we can see a clear gap between
$a_n$
and
$a_{n+1}$
. So, to show such a limit cannot exist, we pick a small
$\varepsilon$
so
$(-1-\varepsilon, 1+\varepsilon)\cap (1-\varepsilon,1+\varepsilon) = \emptyset.$
That is, the two intervals do not touch (shown by red bars in the visual). A rule of thumb is to pick
$\varepsilon$
to be
$1/4$
the distance between two points, as done here.