Divergent Sequence may be Bounded

Theorem
While every Convergent Sequence is Bounded, it does not follow that every bounded sequence is convergent.

There exist bounded divergent sequences.

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.

Alternative 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 $-1$.

Note the following subsequences of $\left \langle {x_n} \right \rangle$:
 * $\left \langle {x_{n_r}} \right \rangle$ where $\left \langle {n_r} \right \rangle$ is the sequence defined as $n_r = 2r$;
 * $\left \langle {x_{n_s}} \right \rangle$ where $\left \langle {n_s} \right \rangle$ is the sequence defined as $n_s = 2s+1$.

The first is $1, 1, 1, 1, \ldots$ and the second is $-1, -1, -1, -1, \ldots$

So $\left \langle {x_n} \right \rangle$ has two subsequences with different limits.

From Limit of a Subsequence, that means $\left \langle {x_n} \right \rangle$ can not be convergent.