Divergent Sequence may be Bounded/Proof 2

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 $\left \langle {x_n} \right \rangle$:
 * $(1): \quad \left \langle {x_{n_r}} \right \rangle$ where $\left \langle {n_r} \right \rangle$ is the sequence defined as $n_r = 2r$
 * $(2): \quad \left \langle {x_{n_s}} \right \rangle$ 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 Subsequence equals Limit of Sequence, that means $\left \langle {x_n} \right \rangle$ can not be convergent.