$\sup$ and $\inf$ of this set These are exercises from my textbook, and I am not sure if the solutions are correct or not.
Given a set $B = \{\frac{n}{2n+1} : n \in \mathbb{N} \}$
Find the $\sup$ and $\inf$ of $B$, and maxima and minima if they exist.
The solutions give $\sup B = \frac{1}{2}$ and $\inf B = \frac{1}{3}$
I am confused by this result. The $\sup$ has to be the least upper bound, $\frac{1}{2} \in B$ and $\frac{1}{3}$ is also in $B$. So how can we claim that $\sup B = \frac{1}{2}$ when $\frac{1}{3} > \frac{1}{2}$
Is this a typo?
Additionally, shouldn't $\inf B = \frac{1}{2}$ since as $n \to \infty$, $\frac{n}{2n+1} \to \frac{1}{2}$
 A: Let's see whether
$$
\frac{n}{2n+1}<\frac{n+1}{2(n+1)+1}
$$
which is equivalent to
$$
n(2n+3)<(n+1)(2n+1)
$$
or
$$
2n^2+3n<2n^2+2n+n+1
$$
that is, $0<1$ which is true.
Therefore the sequence $n/(2n+1)$ is increasing. If you allow $0$ in the natural numbers, the minimum is $0$; otherwise it's $1/3$.
Since for natural $n$ we have $n/(2n+1)<1$, because this is equivalent to $n<2n+1$ which is true, the sequence is bounded and the supremum is the limit.
If you can't apply limits in the proof, use them to find what the supremum must be, that is, $1/2$ and verify by hand it's indeed the supremum.
First:
$$
\frac{n}{2n+1}\le\frac{1}{2}\iff 2n\le 2n+1\iff 0\le1
$$
so indeed $1/2$ is an upper bound. Let $\varepsilon>0$ and try solving
$$
\frac{n}{2n+1}>\frac{1}{2}-\varepsilon
$$
If this has solution for every $\varepsilon$, you're done.

 $\displaystyle\varepsilon>\frac{1}{2}-\frac{n}{2n+1}\iff\varepsilon>\frac{1}{2n+1}\iff 2n+1>\frac{1}{\varepsilon}$

A: We have
$$
0.333... = \frac{1}{3} < \frac{1}{2} = 0.5
$$
so the $\sup$ is right.
Furthermore, $\frac{1}{3} < \frac{1}{2}$ also implies that the infimum is at most $\frac{1}{3}$.
A: First $\frac{1}{3} < \frac{1}{2}$. 
Second $B$ is bounded above by $\lim \frac{n}{2n+1} = \frac{1}{2}$. Then $\sup B \leq \frac{1}{2}$. Take any $\epsilon > 0$ then there exists $N \in \mathbb N$ such that 
$$n > N \implies |x_n - \frac{1}{2}| < \epsilon $$ 
then there exists  $x_k = \frac{k}{2k + 1} \in B$, such that $$\frac{1}{2} - \epsilon < x_k \leq \epsilon$$ then $\sup B = \frac{1}{2}$. 
Now $\inf B = \frac{1}{3}$ because when $n = 1$ you have $\frac{1}{2 \dot \ 1 + 1} = \frac{1}{3}$. Then $B$ is bounded below by $\frac{1}{3}$, try to show that in fact $\inf B = \frac{1}{3} $.  
