Prove $(x_n)$ does not converge to real number $x$ using definition

Question

$x_n$ = $(-1)^n(1-{1\over n})$

Prove: If $x$ ∈ $\Bbb{R}$ is any real number, then ($x_n$) does not converge to $x$.

This has to be proved using the negation of the definition i.e.

$\exists{\varepsilon}>0$, $\forall{N∈\Bbb{N}}$, $\exists{n}>N$ : $|x_n-x|\ge\varepsilon$

Answer 1

Suppose $\lim_{n\to\infty}x_n=x\in \mathbb{R}$ Let $\epsilon=\frac{1}{4}$

Suppose $x\leq 0, \forall n>1,(-1)^{2n}(1-{1\over 2n})>\frac{1}{2}$, Hence $\forall N\in \mathbb{N}$, pick $n=2N>N$, then $$|(-1)^{n}(1-{1\over n})-x|=(-1)^{n}(1-{1\over n})-x\geq(-1)^{n}(1-{1\over n})>\frac{1}{2}>\epsilon$$

Similarly for $x> 0$ by considering $n$ is odd.

By definition, it doesn't converge to any real number.

Answer 2

$\exists{\varepsilon}>0, \forall{N∈\Bbb{N}}, \exists{n}>N : |x_n-x|\ge\varepsilon$

Let's take $\varepsilon = \frac{1}{2}$. $\forall n > 1: |x_n| \geq \frac{1}{2}$ and $|x_n - x_{n+1}| \geq 1$. Then even if for some $n$: $|x_n - x| < \varepsilon$ then $|x_{n+1} - x| = |x_{n+1} - x_n + x_n - x| \geq |x_{n+1} - x_n| - |x_n - x| > 1- \varepsilon$.