prove the limit of sequence $\frac{2n}{n+1} = 2$ Please check the proof and point the mistake:
$$\frac{2n}{n+1}-2< \epsilon $$
$$\frac{2n-2n+2}{n+1}< \epsilon $$
$$\frac{2}{n+1}< \epsilon$$ 
Note: If we end the proof at this point I think the proof might fail if $n=1$.
I'm not sure about it.
$$2\left | \frac{1}{n+1} \right |< \epsilon $$
$$\left | \frac{1}{n+1} \right |< \frac{\epsilon }{2}$$
Then
$$\frac{2}{n+1}=2\left | \frac{1}{n+1} \right |$$
$$2\left | \frac{\epsilon }{2} \right |=\epsilon $$
 A: In the second step, I think it's possible you didn't distribute the minus sign. Shouldn't it be $2n-2n-2$, not $2n-2n+2$, in the numerator? I'm not sure that this is considered rigorous, but you could simply factor out $n$ from both the numerator and denominator of the original expression. Those $n$s would then cancel and you'd be left with $\frac{2}{1 + \frac{1}{n}}$, which is a much easier limit to evaluate.
A: Just as an example this is what I would do: if our assumption is true (i.e., if $2$ is the limit of the sequence) then following the definition of limit we know that for any $\epsilon>0$ exists some $N\in\Bbb N$ such that
$$\left|\frac{2n}{n+1}-2\right|<\epsilon\implies\left|\frac{2n-2n-2}{n+1}\right|<\epsilon\implies \frac2{n+1}<\epsilon,\quad\forall n\ge N$$
Then it is enough to observe that the RHS is arbitrarily small, by example choose $N\ge\frac2{\epsilon}$ (by the archimedean property of the reals this is possible), then we will had
$$1<\frac{\epsilon}2\left(\frac2{\epsilon}+1\right)=1+\frac{\epsilon}2\le N\frac{\epsilon}2+\frac{\epsilon}2$$
what is true, because $\epsilon>0$ and for the chosen $N$ we have that $N\frac{\epsilon}2\ge 1$.

A different approach, maybe more easy, just observe that $\frac1n>\frac1{n+1}$ ( let $n\ge 1$). Then because you know that the sequence $(1/n)$ converges to zero then it is true that it does for $(1/(n+1))$, because the first is bigger than the second.
And observe that $\epsilon/2<\epsilon$, so it doenst change the truth of the inequality. If you have that 
$$\left(\frac1n<\epsilon\right)\land\left(\frac1{n+1}<\frac{\epsilon}2<\epsilon\right)\land\left(\frac1{n+1}<\frac1n\right)\implies \frac1{n+1}<\frac1n<\epsilon,\quad\forall n\ge N>\frac1\epsilon$$
and then $(1/(n+1))$ converges to zero.
A: Consider the corresponding real function over the positive reals. The derivative is positive by the quotient rule and the closure of the positive reals. The function never assumes 2. It remains only to show it assumes all values slightly less than 2: Solve (2x)/(x+1)=2-2e . That is x/(x+1)=1-e, or x=x-xe+(1-e). This is a linear equation and thus has a solution as desired.
