Fake proof for a sequence convergence to a limit

Where does a false proof of a sequence breaks down. For example the sequence:
$$\lim_{} \left(\frac{n+1}{n}\right)=1$$ If I want to prove that a sequence converges to a limit other than 1, take $0$.

Is it the absolute value that is incorrect, in that case why? $$\left\vert \frac{n+1}{n} -0 \right \vert =\left\vert \frac{n+1}{n} \right \vert = \frac{n+1}{n} $$ $$ \frac{n+1}{n}<\epsilon \Rightarrow n+1<\epsilon n \Rightarrow1 <\epsilon n-n \Rightarrow 1 <n( \epsilon-1) \Rightarrow \frac{1}{\epsilon-1}<n $$ This is a question from the book understanding analysis by Stephen Abbott that a really can't get my head around.

Your proof is wrong because It does not work for $\epsilon$ less than 1. Since $\epsilon$ is required to be taken as small as possible, you could assume without to lose the generality that all $\epsilon$ in all proofs are less than one. This assumption would make Impossible to make the passage $$ 1< n (\epsilon -1 ) \implies \dfrac{1}{\epsilon -1 } < n. $$ The actual passage is $$ 1< n (\epsilon -1 ) \implies \dfrac{1}{\epsilon -1 } > n$$ because the quantity $(\epsilon -1 )$ is negative which requires changing the sign of less than to bigger than. The actual needed last conclusion is impossible.