Check that that the sequence $(-1)^{n}\left ( 1-\frac{1}{n} \right )$ converge or not. Please check my proof and point the flaw in the proof :)
$(-1)^{n}\left ( 1-\frac{1}{n} \right )&lt; \epsilon $

because n is even the sequence will be

$1\left ( 1-\frac{1}{n} \right )$

$\left ( 1-\frac{1}{n} \right )&lt; \epsilon $

$\frac{n-1}{n}&lt; \epsilon $

$n-1&lt; \epsilon n$

since inequality hold Limit exist and convergence to some number.

we conclude for $|\frac{n-1}{n}|&lt; \epsilon $

If n is odd

$(-1)^{2n+1} = -1$

$ -1\left ( 1-\frac{1}{n} \right )&lt; \epsilon $

$ -1+\frac{1}{n}&lt; \epsilon $

$ \frac{-n+1}{n}&lt; \epsilon $

because the inequality hold limit exist at some number can converge to the same

then $\frac{-n+1}{n}&lt; \epsilon $

But $|\frac{-n+1}{n}\neq |\frac{n-1}{n}|$

and limit is unique.

We conclude no limit exist for this sequence;divergent
How does the inequality $n-1&lt;\epsilon n$ hold?

For example, what if $\epsilon = 0.001$ and $n=1000$? The inequality then becomes $999 &lt; 1$, something you will probably agree is not true. In fact, if $\epsilon \leq \frac12$, then the inequality
$$n-1&lt;\epsilon n$$

will only hold for $n=1$, since for $n\geq 2$, you have $$\epsilon n &lt; \frac 12 n =n-\frac 12n  \leq n-\frac 12 \cdot 2 = n-1$$

i.e. the oposite of your inequality.



What you need to show is not that $1-\frac 1n$ is small (it is not), but that $1-\frac 1n$ is close to $1$. In other words, you have to show that for every $\epsilon&gt;0$ there exists some $N$ such that if $n&gt;N$, then the distance between $1$ and $1-\frac1n$ is smaller than $\epsilon$ (I'll leave it to you to fully formalize this statement).

That, along with showing that $-(1-\frac1n)$ is close to $-1$, is enough to see that the sequence has two convergence points and thus cannot converge.
Let  $a_n:=(-1)^{n}\left ( 1-\frac{1}{n} \right )$ 

Then $a_{2n}=\left ( 1-\frac{1}{n} \right ) \to 1$ and $a_{2n-1}=\left ( -1+\frac{1}{n} \right )\to -1$