I was looking at an exercise in which I have to find the supremum and infimum of a set , and I need some help with the infumum part. The set is :
$$A = \left\{ \frac{n^2+2n+3}{4n^2+2n} \; \middle| \; n \text{ a natural number (different from 0)} \right\}$$
If we consider the function $f(x) = \frac{x^2 +2x +3}{4x^2 +2x}$ we have a decreasing function from $x>0$ and therefore $f(n)$ is a decreasing function. Since the set $A$ is the range of $f(n)$ for the indicated $n$’s, its clear that the upper bound is $1$, and the lower bound is $1/4$ (the value of the limit of $f(n)$ when $n$ goes to infinity).
So, for the supremum part :
Let $M$ be an upper bound. We want to show that $1\leq\ M$. Suppose $1 \gt\ M$ For def of upper bound we have $f(n) \leq \ M$. So $ f(n) \leq\ M \lt\ 1$. If $n=1$, then $ 1\leq\ M \lt\ 1 $ and this is a contradiction caused by saying that $1 \gt\ M$ , therefore $1 \leq\ M$ and$ 1$ is supremum by definition of supremum.
Infimum proof : Here I don't know how to use the fact that the limit of $f(n)$ when $n$ goes to infinity is $1/4$. Can you give me some ideas please? Thanks!
You have $\sup A=1$ because $1\in A$ and because $a\in A\implies a\leqslant1$ (so, in fact, $1=\max A$).
On the other hand, if $a\in A$, then $a>\frac14$, and therefore $\frac14$ is a lower bound of $A$. Now, take $r>\frac14$. and let $\varepsilon=r-\frac14$. Since $\varepsilon>0$, there is some $N\in\Bbb N$ such that\begin{align}n\geqslant N\implies\left|\frac14-\frac{n^2+2n+3}{4n^2+2n}\right|<\varepsilon&\iff\frac{n^2+2n+3}{4n^2+2n}-\frac14<\varepsilon=r-\frac14\\&\iff\frac{n^2+2n+3}{4n^2+2n}<r,\end{align}and so $\frac{N^2+2N+3}{4N^2+2N}<r$. But $\frac{N^2+2N+3}{4N^2+2N}\in A$. Therefore, it is proved that $r$ is not a lower bound of $A$. It follows from this that $\frac14$ is the greatest lower bound of $A$. In other words, $\inf A=\frac14$.