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Divergent Sequences

Example 1.
Prove the sequence
{an}\{a_n\}
​ given by
an≜(−1)na_n \triangleq (-1)^n
​ for all
n∈Nn\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
ε\varepsilon
​ that would prove, by way of contradiction, the limit definition does not hold?
The sequence oscillates between
−1\mathsf{-1}
and
1\mathsf{1}
. Letting ​
ϵ=1/2\mathsf{\epsilon=1/2}
gives a contradiction.
By way of contradiction, suppose
{an}\{a_n\}
​ converges. This implies there is
α∈R\alpha\in\mathbb{R}
​ and
N∈NN\in\mathbb{N}
​ such that
∣an−α∣≤1/2|a_n - \alpha|\leq 1/2
​ for all
n≥Nn\geq N
​. In particular,
|
12≥∣a2N−α∣=∣1−α∣≥1−α  ⟹  α≥12\dfrac{1}{2} \geq |a_{2N} - \alpha| = |1-\alpha| \geq 1-\alpha \quad \implies\quad \alpha \geq \dfrac{1}{2}
| | | ---------------------------------------------------------------------------------------------------------------- | :-: |
​and
|
12≥∣a2N+1−α∣=∣(−1)−α∣≥−1−α  ⟹  α≤−12.\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
​
12≤α≤−12\dfrac{1}{2} \leq \alpha \leq -\dfrac{1}{2}
​
​
a contradiction. Thus, the initial assumption
{an}\{a_n\}
​ converges was false, and we are done.
​
■\blacksquare
​
​
To tackle this problem, we first label
{an}\{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
α\alpha
. However, here we can see a clear gap between
ana_n
and
an+1a_{n+1}
. So, to show such a limit cannot exist, we pick a small
ε\varepsilon
so
(−1−ε,1+ε)∩(1−ε,1+ε)=∅.(-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
ε\varepsilon
to be
1/41/4
the distance between two points, as done here.