Help with prove of inf.

Question

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!

Answer 1

Let: $$ S = \left\{\frac{1}{3} + \frac{3n + 1}{6n^2} ~\middle|~ n \in \mathbb N\right\} $$ Notice that since $3n + 1 > 0$ and $6n^2 > 0$, we know that $\frac{3n + 1}{6n^2} > 0$ 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 $\epsilon > 0$. Now recall that, by the Archimedean property, there is some $N \in \mathbb N$ such that $N > \frac{2}{3\epsilon}$. But then since $\frac{1}{3} + \frac{3N + 1}{6N^2} \in S$ and: \begin{align*} \frac{1}{3} + \frac{3N + 1}{6N^2} &< \frac{1}{3} + \frac{3N + N}{6N^2} &\text{since }N > 1 \\ &= \frac{1}{3} + \frac{2}{3N} \\ &< \frac{1}{3} + \epsilon &\text{since }N > \frac{2}{3\epsilon} \iff \epsilon > \frac{2}{3N} \\ \end{align*} it follows that $\frac{1}{3} = \inf S$, as desired. $~~\blacksquare$

Answer 2

First note that ${1 \over 3} + { 3n+1 \over 6n^2} \ge {1 \over 3}$, so ${1 \over 3} $ is a lower bound.

Now let $\epsilon >0$ and choose $n$ large enough so that ${ 3n+1 \over 6n^2} < \epsilon$. (One easy way is to choose $n$ large enough so that ${1 \over n } < \epsilon$, then $\epsilon > { 1\over n} = { 6 n \over 6 n^2} > { 4n \over 6 n^2} \ge {3n+1 \over 6 n^2}$.)

Then ${1 \over 3} + { 3n+1 \over 6n^2} < {1 \over 3}+ \epsilon$.

It follows that ${ 1\over 3} = \inf_{n \in \mathbb{N}} ({1 \over 3} + { 3n+1 \over 6n^2} )$.

Answer 3

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 $\inf=\frac{1}{3}$