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I came across an example while studying which asks us to solve (and prove by definition), the supremum and infimum of the following set:

A = {$\frac{n^3+1}{n^4+16}$:$n = 1,2,3...$}

Here is my approach. I would like to know if this works or if I'm getting lost in the objective to prove.


I start by looking for the $\inf(A)$. Clearly the infimum is $0$ due to the dominating exponent in the denominator. So by the definition, I need to show;

  1. $0$ is a lower bound for $A$
  2. $\forall $ $ \epsilon>0$, $0 + \epsilon$ is not a lower bound for A

I begin with a rough proof to find the logic I need to follow for 1. So, show:

\begin{align} 0 \leq \frac{n^3+1}{n^4+16} \\ 0 \leq n^3+1\\ 0 \leq n+1 \leq n^3+1\\ -1 \leq n \end{align}

Which then lets me start the actual proof:

By the Archimedean property, $\exists$ $n \in \mathbb{N}$ s.t. $n \geq -1$

\begin{align} n+1 \geq 0\\ n^3+1 \geq n+1 \geq 0\\ \frac{n^3+1}{n^3+16} \geq 0 \end{align}

So $0$ is a lower bound for A

Secondly, we need to show that $\epsilon + 0$ is not a lower bound for A. So, starting with a rough proof, I need to show that there exists some $n$ s.t

\begin{align} \frac{n^3+1}{n^4+16} \leq \epsilon\\ \frac{n^3+1}{n^4+16} \leq \frac{n^3}{n^4+16}+1 \leq \epsilon\\ \frac{n^3}{n^4+16} \leq \epsilon -1\\ \frac{n^3}{n^4+16} \leq \frac{n^3}{n^4} \leq \epsilon -1\\ \frac{1}{n} \leq \epsilon -1\\ n \geq \frac{1}{\epsilon -1} \end{align}

And with this, I worked backward to get the implication I needed (starting with the A.P as stated above). I won't write out my entire work for the supA, but it follows the same sort of method I used here. I want to make sure these step are coherent, and is there anything I can do to get to the desired values quicker as this is quite a tedious process to do explicitly. Thanks!


Edit:

In my supremum proof I've found an inequality I can't seem to manipulate. After proving 28/97 is an upper bound (through inspection of the sequence, n=3 yields the largest value). In the second step of showing that $\frac{28}{97} -\epsilon$ is not an upper bound, I begin like usual (showing that there exists an n s.t.):

\begin{align} \frac{n^3+1}{n^4+16} \geq \frac{28}{97} - \epsilon\\ \frac{n^3+1}{n^4+16} \geq \frac{n^3}{n^4+16} \geq \frac{28}{97} - \epsilon \end{align}

And I seem to be stuck. No matter the manipulation I can think of, I can't get myself to a form of $n \geq f(\epsilon)$ to use the Archimedean property. Any hints?

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    $\begingroup$ To establish that $0$ is a lower bound note that the positive reals are closed under addition, multiplication, and reciprocation. While you can always use $\leq$ sometimes $<$ is more accurate and should be preferred. For example steps like $n^3 + 1 > n^3$ should use the strict inequality. Also you need to establish a term is $< \epsilon$ not merely $ \leq \epsilon$. Other than that your second part of the proof looks fine. $\endgroup$ Oct 18, 2023 at 2:02
  • $\begingroup$ @CyclotomicField so just to be clear. When proving the seq is less than some value, it suffices to show that a bigger seq is less than that value (since the smaller one with also be). And when showing the seq is bigger than some value, it suffices to show that a smaller seq is bigger than that value (for the same reasons)? $\endgroup$
    – aort01
    Oct 18, 2023 at 2:06
  • $\begingroup$ If you think of it that way it's called the squeeze theorem although most of these follow from properties of an ordered field so you don't need that much. en.wikipedia.org/wiki/Squeeze_theorem $\endgroup$ Oct 18, 2023 at 2:10
  • $\begingroup$ @CyclotomicField I've attached a comment for a problem I'm having with the sup proof. Is there any manipulation to get me in the form of the A.P? Or another theorem I can start off with when working backwards? $\endgroup$
    – aort01
    Oct 18, 2023 at 2:27
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    $\begingroup$ @Alex03 Seems that your reasoning on $$\frac{1}{n} \leq \varepsilon -1 \iff n \geq \frac{1}{\varepsilon -1}$$ is mistaken. Note that $\varepsilon$ can be any positive number. Take $\epsilon = 0.5$, your assertion amounts to $$\frac{1}{n} \leq -0.5 \iff n \geq -2 $$ which is clearly wrong. $\endgroup$ Oct 18, 2023 at 6:51

1 Answer 1

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My reasoning is too long to write in the comment. So I write my own answer here.

$(1)$ To show that $0$ is the infimum.

$(i)$ $n^3+1 \gt 0$ and $n^4+16 \gt 0 \implies \frac{n^3+1}{n^4+16} \gt 0$.

$\therefore 0$ is a lower bound.

$(ii)$ For any $\varepsilon \gt 0$, we reason as follows:

$$\frac{n^3+1}{n^4+16} \lt \frac{n^3+n^3}{n^4}=\frac{2}{n}$$

So if we can find $n$ such that $\frac{2}{n} \lt \varepsilon$, then we are done. But this is easy, for this is equivalent to $n \gt \frac{2}{\varepsilon}.$

Formally we can write the proof as follows:

For any $\varepsilon \gt 0$, take $n_0=\lfloor \frac{2}{\varepsilon} \rfloor +1 $, then $n_0 \gt \frac{2}{\varepsilon}$ and hence $\varepsilon \gt \frac{2}{n_0}$.

Therefore $$\frac{n_0^3+1}{n_0^4+16} \lt \frac{n_0^3+n_0^3}{n_0^4}=\frac{2}{n_0} \lt \varepsilon$$

$(i)$ and $(ii)$ prove that $0$ is the infimum. $$$$ $(2)$ To prove that $\frac{28}{97}$ is the supremum, we prove that $\frac{28}{97}$ is one of the terms and $$\frac{n^3+1}{n^4+16} \leq \frac{28}{97} \;\;\; \forall n \in \mathbb N.$$

Let $a_n=\frac{n^3+1}{n^4+16}$ and put $n=1, 2, 3, 4,$ we find that $$a_1=\frac{2}{17}, a_2=\frac{9}{32}, a_3=\frac{28}{97}, a_4=\frac{65}{272}.$$ We see that $a_3=\frac{28}{97}$ is the largest of the the first $4$ terms.

Next we prove that $$\frac{n^3+1}{n^4+16} \leq\frac{28}{97} \;\; \forall n \geq 5. $$

This can be done as follows:

$\forall n \geq 5,$ \begin{align} \frac{n^3+1}{n^4+16} & \lt \frac{n^3+1}{n^4} \\ & = \frac{1}{n}+\frac{1}{n^4} \\ & \leq \frac{1}{5}+\frac{1}{5^4} \\ & = \frac{126}{625} \\ & \lt \frac{28}{97} \end{align}

This completes the proof that $\frac{28}{97}$ is the supermum.

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  • $\begingroup$ That is a great help in solidifying the proof. Thank you very much $\endgroup$
    – aort01
    Oct 18, 2023 at 12:57

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