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.
 A: You have already shown that your $N$ works, I think it's all the notation that is giving you a headache. I would write like this (these are your words, but rewritten):
Given $\varepsilon$, take any $N > \frac{3}{2\epsilon}$. Now, for any $n >N$, we have $$ \frac{3}{2n} < \varepsilon$$
and so, 
$$\left|\frac{2+\frac{3}n}{2+\frac{1}n} - 1\right| < \left|1 + \frac{3}{2n} - 1\right| = \frac{3}{2n} < \varepsilon $$
A: It might be useful to rewrite the expression in question before doing "epsilontic":
$$u_n = \frac{2n+3}{2n+1} = \frac{2n+1+2}{2n+1}= 1+\frac{2}{2n+1}$$
Now,
$$u_n - 1 = \frac{2}{2n+1} \lt \frac{2}{2n}= \frac{1}{n}$$
From here. you can surely find your $N_{\varepsilon}$ easily.
A: The answer of @JuliusL33t is right on the nose. 
But I want to show how you could have arranged your work and derived a value for $N$ in a way that allows you to close off the proof more smoothly (this will be a different value of $N$).
The idea is to work backwards from the inequality $|u_n-1| < \epsilon$, rewriting it in equivalent form as you go along (if that is possible, which for this problem it is).
Start with the inequality
$$(1) \qquad |u_n - 1| < \epsilon
$$
By substitution, inequality (1) is logically equivalent to
$$(2) \qquad \bigg| \frac{2n+3}{2n+1} - 1 \bigg| < \epsilon
$$
By the laws of algebra, inequality (2) is logically equivalent to
$$(3) \qquad \bigg| \frac{2}{2n+1} \biggr| < \epsilon
$$
Since $n$ is a positive integer, the fraction inside the absolute value signs is positive, and so we may strip off the absolute value signs, to get a logically equivalent inequality
$$(4) \qquad \frac{2}{2n+1} < \epsilon
$$
By the laws of inequality, (4) is logically equivalent to
$$(5) \qquad n > \frac{2-\epsilon}{2\epsilon}
$$
Now take any $N > \frac{2-\epsilon}{2\epsilon}$. If $n > N$ then (5) is true, and so each of the logically equivalent inequalities (4), (3), (2), (1) is true, hence $|u_n-1|<\epsilon$ is true.
A: 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 $N$
