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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 $$\inf S=\lim_{n}(-1+\frac{19}{2n+1})=-1$$

$$\sup S=-1+\frac{19}{2\cdot 1+1}=\frac{16}{3}$$

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    You're right about sup, you have only to show that -1 is inf. – 2017-01-10
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    @arberavdullahu Didn't I show that? – 2017-01-10
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    You showed that -1 is an lower bound for S, to show that -1 is infimum you have to show that for any r>-1 there exist an x element of S such that x – 2017-01-10
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    Yes, you did. As you found a sequence in $S$ converging to the candidate for the infimum, it obviously respect the condition to be an infimum: for every $\epsilon>0$, there exists $N_{\epsilon}$ such that $s_{N_{\epsilon}}-\epsilon < -1$. Since $s_{n}\geq -1$ for all $n$, this is equivalent to the fact that $s_{n} converges to $-1$. – 2017-01-10

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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 $\inf\limits_{n\in\mathbb N}f(n)=\lim\limits_{n\rightarrow+\infty}f(n)=-1$.