Exercises: Convergence and divergence – "Math for Non-Geeks"

Proving divergence for an alternating sequence

Proof (Alternative proof by contradiction)

For two subsequence elements $a_{n}$ and $a_{n+1}$ there is always $|a_{n}-a_{n+1}|=|(-1)^{n}-(-1)^{n+1}|=|\pm 2|=2.$. So if for $\epsilon =1>0$ there was an $N\in \mathbb {N}$ and an $a\in \mathbb {R}$ with $|(-1)^{n}-a|<\epsilon \ \forall n\geq N$ then, we would have $2=|(-1)^{N}-(-1)^{N+1}|=|(-1)^{N}-a+a-(-1)^{N+1}|{\underset {\text{inequality}}{\overset {\text{triangle-}}{\leq }}}|(-1)^{N}-a|+|a-(-1)^{N+1}|=|(-1)^{N}-a|+|(-1)^{N}-a|<\epsilon +\epsilon =1+1=2$ ↯