Show $S_n = \frac{n}{2n+1}$ converges to $\frac{1}{2}$ This is my thinking so far but it is by no means a rigorous proof:
$\frac{n}{2n+1} < \frac{n}{2n} = \frac{1}{2}$ 
I don't know how to translate this result into a proof, thank you.
 A: $\frac{n}{2n+1}=\frac{n}{n(2+\frac{1}{n})}=\frac{1}{2+\frac{1}{n}}\to\frac{1}{2}$
A: If you want a rigorous proof with all the $\varepsilon$'s then note that
$$
S_n=\frac n{2n+1} = 1 - \frac{n+1}{2n+1} > \frac 12 - \frac 1{2n+1} > \frac 12 - \frac1n
$$
Now you have that 
$$
\frac 12 > S_n > \frac 12 - \frac 1n,
$$
so 
$$
\left|S_n - \frac 12\right| < \frac 1n
$$
Now by the Archimedean property, for all $\varepsilon > 0$ there exists $N\in\mathbb{N}$ such that for all $n> N$, $\frac 1n < \varepsilon$. Thus $S_n\to \frac 12$.
A: The expression $\displaystyle\lim_{n\to\infty}a_n=\alpha$ means that there always exists an integer $N$ for every positive number $\epsilon$ such that $\lvert a_n-\alpha\rvert$ is smaller than $\epsilon$ if $n$ is bigger than $N$. In symbolic words,
\begin{equation}
(\forall\epsilon>0)(\exists N\in\mathbf{N})(\forall n\in\mathbf{N})(n>N\implies\lvert a_n-\alpha\rvert<\epsilon).
\end{equation}
Following this definition, let's see whether $S_n$ converges and indentify what the value is if it does. Let $\epsilon$ be a sufficiently small but positive number and $n>\dfrac{1}{4}\left(\dfrac{1}{\epsilon}-2\right)$. From this inequality, we easily get
\begin{equation}
\frac{1}{2}-S_n<\epsilon.
\end{equation}
Since $S_n$ is always smaller than $\dfrac{1}{2}$ as you mentioned, we get
\begin{gather}
\frac{1}{2}-S_n<\epsilon\ \land\ S_n-\frac{1}{2}<(0<)\epsilon, \\
\therefore\quad\frac{1}{2}-\epsilon<S_n<\frac{1}{2}+\epsilon.
\end{gather}
This finally implies
\begin{equation}
\left\lvert S_n-\frac{1}{2}\right\rvert<\epsilon\quad \text{for any}\ n>\frac{1}{4}\left(\frac{1}{\epsilon}-2\right).
\end{equation}
Thus, $\displaystyle\lim_{n\to\infty}S_n=\dfrac{1}{2}$ (and $N=\left\lfloor\dfrac{1}{4}\left(\dfrac{1}{\epsilon}-2\right)\right\rfloor+1$).
