Prove that the supremum of this set is $\frac{1}{4}$: $S=\{\frac{n-1}{4n+1}: n \in \mathbb{N}\}$ without using limits?

Question

I’m trying to prove that the supremum of this set is $\frac{1}{4}$: $S=\{\frac{n-1}{4n+1}: n \in \mathbb{N}\}$, without using limits.

I can easily show that $\frac{1}{4}$ is an upper bound, but I have trouble showing that it is the least upper bound.

Strategy: 1) show it’s an upper bound by contradiction. 2) then, show that if there existed a lower bound $k$ that was strictly less than $\frac{1}{4}$, then it wouldn’t be an upper bound because we could find an $n$ such that $\frac{n-1}{4n+1}$ was greater than that upper bound. Do this by setting $\frac{n-1}{4n+1}=1/4$, and then solving for $n$, as a function of $k$ and then incrementing it by one, and then showing that if we let $n$ = the function of $k$ incremented by one, then it would be greater than the upper bound $k$.

Proof: Assume for contradiction that $\frac{1}{4}$ is not an upper bound, then there exists $n$ such that $\frac{n-1}{4n+1} > \frac{1}{4}$, then since $4n+1 > 0$, $-4>1$ which is a contradiction. So it is an upper bound. Assume for contradiction that there exists $k$ such that $k < \frac{1}{4}$ and $k$ is an upper bound for $S$. Then since it’s an upper bound, $\frac{n-1}{4n+1} \le k$, which means $n \le \frac{k+1}{1-4 k }$, since $1-4k \ge 0$.

Now I’m not sure where to go from here. I really want to use the Archimedean property, but since I need $n$ to be in $\mathbb{N}$, I’m not sure if it is acceptable to use that. I’m also not sure what I can increment $n$ by as a function of $k$ in order to guarantee that it will still be in $\mathbb{N}$. Can I get some hints maybe?

Answer 1

In (1), the argument by contradiction is unnecessarily complicated. Instead, one could argue more simply that the numerator of $\frac{n-1}{4n+1}$ is $< n$, and its denominator is $> 4n$, so this arbitrary element of $S$ is $< \frac{n}{4n} = \frac{1}{4}$.

In (2), appealing directly to the Archimedean property as you suggest (but also eschewing an argument by contradiction for a second time): given any $k < \frac{1}{4}$, there exists a positive integer $n$ such that $n\left(\frac{1}{4} - k\right) > 1$, i.e. such that $k < \frac{1}{4} - \frac{1}{n} = \frac{n-4}{4n}$. Conveniently, this fraction happens to be less than the element $\frac{n-1}{4n+1}$ of $S$ (indeed, it is less than $\frac{n-3}{4n+1}$, by the result that $\frac{a}{b} < \frac{a+1}{b+1}$ if $a < b$ and $b > 0$), so $k$ is not an upper bound of $S$.

Answer 2

Note that$$\frac14-\frac{n-1}{4n+1}=\frac5{4(4n+1)}$$and$$\frac5{4(4n+1)}<\varepsilon\iff n>\frac{5-4\varepsilon}{16\varepsilon}.$$

Answer 3

Well, if $s = \frac {n-1}{4n+1} \in S; n\in \mathbb N$ then $\frac {n-1}{4n+1} = \frac {n+\frac 14}{4n+1} - \frac {\frac 54}{4n+1} = \frac 14 - \frac {5}{16n + 4}< \frac 14$.

So $\frac 14$ is an upperbound of $S$.

If $k < \frac 14$ and if $\epsilon = \frac 14 -k>0$.

$\frac 5{16n + 4} < \epsilon \iff$

$\frac 5{\epsilon} < 16n + 4 \iff$

$n > \frac {\frac 5\epsilon -4}{16}$.

Such an $n$ can always be found and if we let $n > \frac {\frac 5\epsilon -4}{16}$ we will have:

$k = \frac 14 - \epsilon < \frac 14 - \frac 5{16n+4} =\frac {n-1}{4n+1}\in S$.

So $k$ is not an upper bound of $S$.

And that is the definition of least upper bound. So $\frac 14 =\sup S$.