Divergent Sequence may be Bounded/Proof 2

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 $-1$.

Note the following subsequences of $\sequence {x_n}$:
 * $(1): \quad \sequence {x_{n_r} }$ where $\sequence {n_r}$ is the integer sequence defined as $n_r = 2 r$
 * $(2): \quad \sequence {x_{n_s} }$ where $\sequence {n_s}$ is the integer sequence defined as $n_s = 2 s + 1$.

We have that:
 * $\sequence {x_{n_r} }$ is the sequence $1, 1, 1, 1, \ldots$
 * $\sequence {x_{n_s} }$ is the sequence $-1, -1, -1, -1, \ldots$

So $\sequence {x_n}$ has two subsequences with different limits.

From Limit of Subsequence equals Limit of Real Sequence, that means $\sequence {x_n}$ can not be convergent.