Find the supremum of the following set

Question

Find the supremum of the following set:

$$A:=\left\{{(n-1)\over(2n+3)} : n \in\mathbb{N}\right\}$$

So I have as my answer that $\sup(A) = \frac12$ but need to justify. We were taught to justify in two steps, first show our answer is an upper bound for the set then show it is a least upper bound.

For step $1$: if $a$ is a member of $A$ then $a = (n-1)/(2n+3)$ for some natural number $n$ and $\frac{n-1}{2n+3} \leq \frac{n}{2n} = \frac12$ so $\frac12$ is an upper bound for the set

Step 2 is where I am uncertain. I think I need to show that for any $y<\frac12$ there exists a member of the set between $y$ and $\frac12$ but not sure how to do this. Thanks for all the answers so far!

Answer 1

Hint:

$$\frac{n-1}{2n+3}=\frac12 \frac{2n-2}{2n+3}=\frac{1}{2}\left(1-\frac{5}{2n+3} \right)$$

Response to your edit in your question:

Suppose you are given $y<\frac12$, then $1-2y>0$.

Let $$n>\frac12 \left(\frac{5}{1-2y} -3\right)$$

Then we have

$$2n+3> \frac{5}{1-2y} $$

$$1-2y> \frac{5}{2n+3} $$

$$2y<1-\frac{5}{2n+3} $$

$$y<\frac{1}{2}\left(1-\frac{5}{2n+3}\right)<\frac12 $$

Answer 2

You should rewrite the $$\frac{n-1}{2n+3}$$ as $$\frac12+(\text{something of } n)$$ and find the areas where this expression is monotone sequence of $n$. After that you should find maximum. Also you need to check the limit of this sequence.