Help with prove of inf. I need to prove that:

$$\inf\{\frac{1}{3}+\frac{3n+1}{6n^2} \Big| n\in\mathbb N\}=\frac{1}{3}$$

I get stuck with my proof, I'll write it down.

$$n\geq1$$
$$3n\geq3$$
$$3n+1\geq4$$
$$\frac{1}{3}+3n+1\geq4+\frac{1}{3}$$

Now, I'm having a problem with $6n^2$ if I multiply by $6n^2$, I'll get variable in the express of $4+\frac{1}{3}$.

Any ideas?, Thanks!
Let:
$$
S = 
\left\{\frac{1}{3} + \frac{3n + 1}{6n^2} ~\middle|~ n \in \mathbb N\right\}
$$
Notice that since $3n + 1 &gt; 0$ and $6n^2 &gt; 0$, we know that $\frac{3n + 1}{6n^2} &gt; 0$ so that $\frac{1}{3}$ is a lower bound for $S$. It remains to show that $\frac{1}{3}$ is the greatest lower bound.

To this end, choose any $\epsilon &gt; 0$. Now recall that, by the Archimedean property, there is some $N \in \mathbb N$ such that $N &gt; \frac{2}{3\epsilon}$. But then since $\frac{1}{3} + \frac{3N + 1}{6N^2} \in S$ and:
\begin{align*}
\frac{1}{3} + \frac{3N + 1}{6N^2}
&amp;&lt; \frac{1}{3} + \frac{3N + N}{6N^2} &amp;\text{since }N &gt; 1 \\
&amp;= \frac{1}{3} + \frac{2}{3N} \\
&amp;&lt; \frac{1}{3} + \epsilon &amp;\text{since }N &gt; \frac{2}{3\epsilon} \iff \epsilon &gt; \frac{2}{3N} \\
\end{align*}
it follows that $\frac{1}{3} = \inf S$, as desired. $~~\blacksquare$
Obviously $\frac{1}{3}+\frac{3n+1}{n^2}\geq\frac{1}{3}.$ The sequence $a_n=\frac{1}{3}+\frac{3n+1}{n^2}$ converges to $\frac{1}{3}$ since $\frac{3n+1}{n^2}$ behaves like $\frac{1}{n}.$ Thus $a_n\rightarrow \frac{1}{3}.$ whcih is to say $\inf=\frac{1}{3}$