TITLE: My proof of divergence of $(-1)^n$
QUESTION [8 upvotes]: Should I shorten my proof?
  (Also, should I try to prove without contradiction?)
We consider the sequence
  $(x_n)_{n \in \mathbb{N}}$,
  where
  $x_n = (-1)^n$.
$\textbf{Lemma.}$
  For every element $x_n$ of the sequence $(x_n)$,
  we have $|x_n| = 1$.
  (We could prove this by induction on $n$.)
$\textbf{Theorem.}$
  $(x_n)$ diverges.
$\textit{Proof.}$
  We prove the theorem by contradiction.
  To that end,
  we assume that $(x_n)$ is not divergent, i.e. we assume that it is convergent.
  With that said, we are done as soon as a contradiction is deduced.
  By assumption, there is an $x \in \mathbb{R}$ such that
  \begin{equation*}
    \forall \varepsilon \in \mathbb{R}, \varepsilon > 0 :
    \exists N           \in \mathbb{N}                  :
    \forall n           \in \mathbb{N}, n > N           :
    |x_n - x| < \varepsilon .
  \end{equation*}
  We choose $\varepsilon = 1$.
  By assumption, there is an $N \in \mathbb{N}$ such that
  \begin{equation*}
    \forall n \in \mathbb{N}, n > N :
    |x_n - x| < 1 .
  \end{equation*}
  We choose $n = N + 1$.
  Hence, both $|x_n - x| < 1$ and $|x_{n + 1} - x| < 1$.
  Thus,
  \begin{equation*}
    |x_{n + 1} - x| + |x_n - x| < 2 .
  \end{equation*}
  Moreover,
  \begin{equation*}
    \begin{split}
      2 & =   |2| \\
        & =   |2| \cdot  1  \\
        & =   |2| \cdot |x_{n + 1}| && | \text{ by Lemma} \\
        & =   |2 x_{n + 1}| && | \text{ by multiplicativeness of abs. val.} \\
        & =   |x_{n + 1} + x_{n + 1}| \\
        & =   |x_{n + 1} + (-1)x_{n}| \\
        & =   |x_{n + 1} - x_{n}| \\
        & =   |x_{n + 1} + 0 - x_{n}| \\
        & =   |x_{n + 1} + (-x + x) - x_{n}| \\
        & =   |(x_{n + 1} - x) + (x - x_{n})| \\
        & \le |x_{n + 1} - x| + |x - x_{n}| && | \text{ by subadditivity of abs. val.} \\
        & =   |x_{n + 1} - x| + |x_{n} - x| \qquad && | \text{ by evenness of abs. val.} \\
    \end{split}
  \end{equation*}
  Hence, by transitivity, we have $2 < 2$.
  Obviously, we deduced a contradiction. QED

REPLY [2 votes]: We have
$$ x_{2n} = (-1)^{2n} = 1 \to 1,$$
and
$$ x_{2n+1} = (-1)^{2n+1} = (-1)(-1)^{2n} = -1 \to -1.$$
Therefore, $x_n$ is divergent.