Definitions of sequences and convergence: here, Definitions of subsequences: here.
For the “show the limit” template: here.
Solution
We claim that \(\{a_n\}\) does not converge. To see this, suppose for the sake of contradiction that it does converge to some \(L\). Pick \(\epsilon = \frac{1}{2}\). Since it converges, then there must be an \(N \in \mathbb{N}\) such that when \(n \geq N\), we must have
Observe now that when \(n\) is odd, then \(n=2k+1\) for some \(k\), we see that
as \(k\) grows without bound, the \(\{a_{2k+1}\}\) converges to \(-1\). On the other hand, when \(n\) is even, then \(n=2k\) for some \(k\). Then
as \(k\) grows without bound, the sequence \(\{a_{2k}\}\) converges to \(1\). When \(n \geq N\), both sequences must satisfy
For \(a_{2k}\), this is
and for \(a_{2k+1}\), this is
However, from this we see that no \(L\) can satisfy both. This is a contradiction. Therefore, the sequence doesn’t converge. \(\blacksquare\)
References