Using the $N - \varepsilon$ definition to find the limit of a sequence Yesterday I had a post about this, and it cleared a lot up; however, even though I feel like I understand how to go about solving problems like this, I don't seem to get the right answers. For example:
$u_n = \frac{2n+3}{2n+1}$. We know that this sequence converges to 1, but I want to prove it using the definition.
So I want to show that , $\forall \varepsilon > 0$, $\exists N$ where,$ n > N \implies u_n - 1 < \varepsilon$.
We can manipulate the expression such that $u_n = \frac{2n+3}{2n+1} =\frac{n}n \frac{2+\frac{3}n}{2+\frac{1}n} = \frac{2+\frac{3}n}{2+\frac{1}n}$
Thus, we can show that $\forall \varepsilon > 0$, $\exists N$ where,$ n > N \implies \frac{2+\frac{3}n}{2+\frac{1}n} - 1 < \varepsilon$
Now, we can say that 0 < $\frac{2+\frac{3}n}{2+\frac{1}n}$ < $\frac{2+\frac{3}n}{2} = 1 + \frac{3}{2n}$.
We can use this logic to say that 
$\frac{2+\frac{3}n}{2+\frac{1}n} - 1 < 1 + \frac{3}{2n} - 1 < \varepsilon$ (since $1 + \frac{3}{2n}$ converges to 1).
So, finding an $N$ for $\frac{3}{2n} - 1$ would be the same as finding an $N$ for $u_n$, so we do:
$1 + \frac{3}{2n} - 1 < \varepsilon$ is the same as $\frac{3}{2n} < \varepsilon$.
Thus, if $ N > \frac{3}{2\varepsilon}$, then the initial implication follows. However, I'm not sure how to complete things from here; am I close, or did I fall off the rails at some point? Thanks.

I think you are giving yourself too much work here.
Straightforwardly $u_n-1=\frac {2}{2n+1}$ and if $\frac 2{2n+1} \lt \epsilon$ we have $n\gt  \frac 1{\epsilon}-\frac 12$
This is reversible, so this will find you a candidate for $N$
