Example
Prove that \(((-1)^n)_n\) doesn't converge to \(1\).

Strategy/Notes

If we write the negation of the limit defintion we get: \(\lim_{n \rightarrow \infty} x_n \neq x\) if there exists an \(\epsilon_0 > 0\) such that for all choices of \(N \in \mathbb{N}\), we can find an \(n \geq N\) such that

$$ \begin{align*} |x_n - x| \geq \epsilon_0. \end{align*} $$

So this shows that the limit doesn’t exist for a specific given value \(x\). This doesn’t show divergence in general. To show that a sequence divergence using this, we would need to show this is true for all choices of \(x\).

But luckily here, we want to show that the limit isn’t \(1\) so we can use the negation of the definition of limit. How do we pick \(\epsilon_0\)? The sequence is \((-1,1,-1,1,-1,1,-1,\ldots)\). Even after a large \(N\), we still alternate between \(-1\) and \(1\). So we want \(\epsilon_0\) to be larger than the distance between \(-1/1\) and \(x=1\). Specifically if let \(\epsilon_0 = 2\), then no matter what \(N \in \mathbb{N}\) we choose, we can always find some \(n \geq N\) (For example, take \(n = 2N + 1\)), such that \(|(-1)^n - 1| = 2 \geq 2\).

Proof

Let \(x_n = (-1)^n\). Let \(\epsilon_0 = 2\). Let \(N \in \mathbb{N}\) be given. Choose \(n = 2N + 1\). Then, \(n \geq N\) and

$$ \begin{align*} |x_n - 1| = |(-1)^{2N+1} - 1 | = |-1 -1| = 2 \geq \epsilon_0. \end{align*} $$

Therefore, \(\lim_{n \rightarrow \infty} x_n \neq 1\). \(\ \blacksquare\)

Example
Prove that \((-1^n)_n\) diverges.

Proof

Consider the subsequence

$$ \begin{align*} (-1^{2k})_k = (1,1,1,1,1,1,...) \end{align*} $$

Clearly this subsequence converges to \(1\). Now, consider the subsequence

$$ \begin{align*} (-1^{2k+1})_k = (-1,-1,-1,-1,-1,-1,...) \end{align*} $$

Clearly, this subsequence converges to \(-1\). Therefore, since \((-1^n)_n\) has two subsequences that converge to two different limits, then \((-1^n)_n\) diverges by this theorem. \(\ \blacksquare\)

References