Find the supremum of the following set

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.

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

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

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

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

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