Divergent Sequence may be Bounded/Proof 1

Proof
Let $\sequence {x_n}$ be the sequence in $\R$ defined as:
 * $x_n = \paren {-1}^n$

It is clear that $\sequence {x_n}$ is bounded: above by $1$ and below by $-1$.

$x_n \to l$ as $n \to \infty$.

Let $\epsilon > 0$.

Then $\exists N \in \R: \forall n > N: \size {\paren {-1}^n - l} < \epsilon$.

But there are values of $n > N$ for which $\paren {-1}^n = \pm 1$.

It follows that $\size {1 - l} < \epsilon$ and $\size {-1 - l} < \epsilon$.

From the Triangle Inequality for Real Numbers, we have:

This is a contradiction whenever $\epsilon < 1$.

Thus $\sequence {x_n}$ has no limit and, while definitely bounded, is unmistakably divergent.