Find the infimum and supremum (if they exist) of the following set:
$$S=\{\frac{18-2n}{2n+1}: n\in\Bbb N\}$$ I'm not sure whether my solution is correct:
$$\frac{18-2n}{2n+1}=-1+\frac{19}{2n+1}>-1$$
$S$ is a decreasing sequence $\Leftrightarrow$ $a_n>a_{n+1}\Leftrightarrow \frac{19}{2n+1}>\frac{19}{2n+3}\Leftrightarrow 2n+1<2n+3 \Leftrightarrow 1<3$
$S$ is decreasing and bounded from below so it's convergent, and $$\inf S=\lim_{n}(-1+\frac{19}{2n+1})=-1$$
$$\sup S=-1+\frac{19}{2\cdot 1+1}=\frac{16}{3}$$
Your solution is correct.
Let $f(n)=-1+\frac{19}{2n+1}$.
Since $f$ is a decreasing function, we obtain:
$\sup\limits_{n\in\mathbb N}f(n)=f(1)=\frac{16}{3}$ and $\inf\limits_{n\in\mathbb N}f(n)=\lim\limits_{n\rightarrow+\infty}f(n)=-1$.