Divergent Sequences

Example 1.
Prove the sequence {an}\{a_n\}{an​}
​ given by an≜(−1)na_n \triangleq (-1)^nan​≜(−1)n
​ for all n∈Nn\in\mathbb{N}n∈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\varepsilonε
​ that would prove, by way of contradiction, the limit definition does not hold?
The sequence oscillates between −1\mathsf{-1}−1
 and 1\mathsf{1}1
. Letting ​ϵ=1/2\mathsf{\epsilon=1/2}ϵ=1/2
 gives a contradiction.
By way of contradiction, suppose {an}\{a_n\}{an​}
​ converges. This implies there is α∈R\alpha\in\mathbb{R}α∈R
​ and N∈NN\in\mathbb{N}N∈N
​ such that ∣an−α∣≤1/2|a_n - \alpha|\leq 1/2∣an​−α∣≤1/2
​ for all n≥Nn\geq Nn≥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}21​≥∣a2N​−α∣=∣1−α∣≥1−α⟹α≥21​
 | | | ---------------------------------------------------------------------------------------------------------------- | :-: |
​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}.21​≥∣a2N+1​−α∣=∣(−1)−α∣≥−1−α⟹α≤−21​.
 | | | ------------------------------------------------------------------------------------------------------------------------ | :-: |
Combining these inequalities reveals
​12≤α≤−12\dfrac{1}{2} \leq \alpha \leq -\dfrac{1}{2}21​≤α≤−21​
​
​
a contradiction. Thus, the initial assumption {an}\{a_n\}{an​}
​ converges was false, and we are done.
​■\blacksquare■
​
​
To tackle this problem, we first label {an}\{a_n\}{an​}
, which conveniently only has two values on the number line. If something converges, then iterates eventually get and stay close to some limit α\alphaα
. However, here we can see a clear gap between ana_nan​
 and an+1a_{n+1}an+1​
. So, to show such a limit cannot exist, we pick a small ε\varepsilonε
 so (−1−ε,1+ε)∩(1−ε,1+ε)=∅.(-1-\varepsilon, 1+\varepsilon)\cap (1-\varepsilon,1+\varepsilon) = \emptyset.(−1−ε,1+ε)∩(1−ε,1+ε)=∅.
 That is, the two intervals do not touch (shown by red bars in the visual). A rule of thumb is to pick ε\varepsilonε
 to be 1/41/41/4
 the distance between two points, as done here.
Properties of Limits
Monotone Sequences
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