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;
- $0$ is a lower bound for $A$
- $\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?
Best Answer
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.