$\epsilon-N$ for $\lim_{n\to\infty} \frac{3n+1}{2n+5}=\frac32$. I want to prove using $\epsilon-N$ that $\lim_{n\to\infty} \frac{3n+1}{2n+5}=\frac32$.
Firstly we will find a sufficently large $n\in \Bbb N$:
\begin{align*}
\epsilon\quad\gt&\quad\left|\frac{3n+1}{2n+5} - \frac32 \right|\\
=&\quad \left|\frac{3n+1 -3n-7.5}{2n+5}\right|\\
=&\quad \left|\frac{-13}{4n+10}\right|\\
\implies&\quad \epsilon \gt \frac{13}{4n+10}\\
\implies&\quad \frac1\epsilon \lt \frac{4n+10}{13}\\
\implies&\quad \frac14\left(\frac{13}\epsilon-10\right) \lt n\\
\end{align*}
So $\forall \epsilon\gt 0, \exists N(\epsilon)>\frac14\left(\frac{13}\epsilon-10\right)$ such that $\forall n\in N, n>N$, $a_n=|\frac{3n+1}{2n+5}-\frac32|\lt \epsilon$
Proof: $\left|\frac{3n+1}{2n+5}-\frac{3}{2}\right| \color{red}{\lt} \left|\frac{3N+1}{2N+5}-\frac{3}{2}\right|\lt\epsilon$
Assuming this is even right. Everything feels good except the $\color{red}{\lt}$ in red, and then when I check, since we have 
\begin{align*}
&\left|\frac{3n+1}{2n+5}-\frac{3}{2}\right| \color{red}{\lt}\left|\frac{3N+1}{2N+5}-\frac{3}{2}\right|,\quad n\gt N\\
&\implies \left|\frac{3n+1}{2n+5} - \frac32\right| - \left|\frac{3(n+1)+1}{2(n+1)+5} - \frac{3}{2}\right|\gt 0
\end{align*}

But this fails for all $n$? What did I do wrong?
Also, since $N\in\Bbb N$, we have to choose our $N$ so it is a natural number simultaneous to being greater than $\frac14\left(\frac{13}\epsilon-10\right)$, how do I ensure that subbing in $\frac14\left(\frac{13}\epsilon-10\right)$ is greater than subbing in the natural number that is the ceiling of this?


What is the problem?
$\large\mid{3n+1\over2n+5}-{3\over2}\mid={13\over4n+10}$
$\large\mid{3N+1\over2N+5}-{3\over2}\mid={13\over4N+10}$
$\large\mid{3(n+1)+1\over2(n+1)+5}-{3\over2}\mid={13\over4(n+1)+10}$
So for $n>N$ the second one is obviously larger than the first one and the first one is obviously larger than the third one.
