Infimum of $\frac{6n}{3n-1}$

Question

Can someone tell me is this is correct or what should I do differently:

Find the infimum of $X=\{{\frac{6n}{3n-1}\}}$.

Let $d>0$. We have to show that $\exists n_0\in\mathbb{N}:\frac{6n}{3n-1}<d$.

$\frac{6n}{3n-1}< d\Leftrightarrow 3n-1 > \frac{1}{d}\Leftrightarrow 3n>\frac{1}{d}+1\Leftrightarrow n>\frac{1}{3d}+\frac{1}{3} \Rightarrow n>\frac{1}{3d}$

$\Rightarrow n_0:=\lceil\frac{1}{3d}\rceil\in\mathbb{N}$ so $inf X =0$

Answer 1

Assuming that the domain for $n$ is the natural numbers, you have

$$\frac{6n}{3n-1} >\frac{6n}{3n} = 2.$$

So you know the infimum is at least $2.$ Since the limit of the same fraction is also $2$, you know that the infimum is exactly $2$. To prove this you need to show that for every $\epsilon >0$, there is some $n_0$ such that if $n>n_0$ then $0\leq \frac{6n}{3n-1} -2 <\epsilon.$ You seem to be missing bits from your definition of infimum.

Edit: I think you're trying to use $d$ where I use $\epsilon$, so I'll switch. Suppose $d>0$ is given. Usually, you have to work these proofs backwards and then write them down forwards. You want

$$\frac{6n}{3n-1} -2 = \frac{2}{3n-1} <d. $$

This would happen if $$\frac{3n-1}{2} > \frac{1}{d} . $$

In turn, this would happen if $$3n-1 > \frac{2}{d}.$$

In turn, this would happen if $$3n>\frac{2}{d}+1.$$

In turn, this would happen if $$n > \frac{2}{3d}+\frac{1}{3}.$$

Now you can write the proof: Let $d>0$ be given. Choose $n_0$ a positive integer larger than $\frac{2}{3d}+\frac{1}{3}.$ Then write down the lines above in reverse order to prove that $\frac{6n}{3n-1} -2 < d$, and you have your proof.

Answer 2

The problem is to prove that

$\text{inf}\,\left\{\dfrac{6n}{3n-1}\right\}_{n\in\mathbb{N}}=2 $

As first step compute some values

$3,\;\dfrac{12}{5},\;\dfrac{9}{4},\;\dfrac{24}{11},\;\dfrac{15}{7},\;\dfrac{36}{17},\;\dfrac{21}{10},\;\dfrac{48}{23},\;\dfrac{27}{13},\;\dfrac{60}{29},\ldots$

Then notice that the sequence is decreasing and that values approach to $2$

We can prove that for any $n$ we have

$\dfrac{6n}{3n-1}<2$

indeed $6n<2(3n-1)\to 6n<6n-2$ is true for any $n\in\mathbb{N}$

then suppose that there exists a value $p>2\to p=2+\varepsilon$ such that

$\dfrac{6n}{3n-1}<p\to \dfrac{6n}{3n-1}<2+\varepsilon$

We can find an index $n_0$ such that for $n>n_0$ we have

$\dfrac{6n}{3n-1}<2+\varepsilon$

indeed

$\dfrac{6n}{3n-1}-2-\varepsilon<0$

is verified taking $n_0= \left\lceil \dfrac{\varepsilon+2}{3 \varepsilon}\right\rceil$

and we are done.

Hope this helps