Divergent Sequence may be Bounded/Proof 1

Proof
Let $\left \langle {x_n} \right \rangle$ be the sequence in $\R$ defined as:
 * $x_n = \left({-1}\right)^n$

It is clear that $\left \langle {x_n} \right \rangle$ is bounded: above by $1$ and below by $-1$.

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

Let $\epsilon > 0$.

Then $\exists N \in \R: \forall n > N: \left|{\left({-1}\right)^n - l}\right| < \epsilon$.

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

It follows that $\left|{1 - l}\right| < \epsilon$ and $\left|{-1 - l}\right| < \epsilon$.

From the triangle inequality, we have:
 * $2 = \left|{1 - \left({-1}\right)}\right| \le \left|{1 - l}\right| + \left|{l - \left({-1}\right)}\right| < 2\epsilon$

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

Thus $\left \langle {x_n} \right \rangle$ has no limit and, while definitely bounded, is unmistakably divergent.