Convergence epsilon - check my proof please

Question

Fairly straightforward question (I hope so anyway), would be very grateful if someone could check my proof.

I need to show that $\frac{n+1}{n+3} \to 1 \space \text{as} \space n \to \infty$.

I start by choosing an arbitrary $\epsilon > 0$.

Then I will set $N \in \mathbb{N}$ such that $N \geq \frac{2}{\epsilon} - 1$

Then for some $n \in \mathbb{N}$ such that $n \geq N$, it holds that

$n \geq \frac{2}{\epsilon} - 1$

$| \frac{n+1}{2} | \geq \frac{1}{\epsilon}$

$| \frac{2}{n+1}| \leq \epsilon$

$| \frac{n+3}{n+1} - 1 | \leq \epsilon$ as required.

Edit: I realise that in some cases, convergence is on the open interval, but my lecturer uses closed epsilon neighborhoods in his definition.

Answer 1

The last step is not good, you have <= and then you just turn it into <. The two are not equivalent. Start with < from the very start and all will be fine. Meaning: choose N such that $N > 2/\epsilon - 1$.

Answer 2

Your answer is perfectly correct ...