I know that $\lim \frac{1}{n} = 0$ as n approaches infinity. I proved it using epsilon-N. But if I am to say, I want to prove that this does NOT converge to one, how would I go on about this? Since I want to prove that it does not converge to one, then $|1/n-1| \geq \epsilon$. Now
$$\left|\frac1n -1\right| = \left|\frac{1-n}{n}\right|\geq \epsilon= |1-n|\geq \epsilon |n|$$
this is where I am stuck. How would I prove it?
On
The definition of convergence requires a fact to be true for every positive $\epsilon.$ So in order to disprove convergence you merely need to find one value of $\epsilon$ for which the fact is false.
Setting $\epsilon =2$ won’t work, because once $1/n$ is less than $1$ unit from $0$ it is less than $2$ units from $1.$ But perhaps you can see some $\epsilon$ that might work.
This is actually a good review of logic and negations.
Proof that $x_n \to L$: For all $\epsilon>0$, there exists $N$ such that for all $n\geq N$, we have $|x_n-L|<\epsilon$.
Proof that $x_n \not\to L$: There exists $\epsilon>0$ such that for all $N$, there exists $n\geq N$ such that $|x_n-L|\geq\epsilon$.
In your case, $x_n=\frac{1}{n}$ and $L=1$. What $\epsilon$ works here?