Show that $\{\frac{n^2}{n^2+n}\}$ is convergent 
Use only the Archimedean Property of $\mathbb{R}$ to show  $\{\frac{n^2}{n^2+n}\}$ is convergent


Let $a_n=\{\frac{n^2}{n^2+n}\}=\{\frac{n}{n+1}\}=\{1-\frac{1}{n+1}\}=\{1\}-\{\frac{1}{n+1}\}$.
$\{1\}$ is convergent to $1$. Show $\{\frac{1}{n+1}\}$ is convergent. So let $\epsilon>0$. By the Archimedean Property, there exists an index $N$ such that $1/N<\epsilon$. So for all $n\geq N$, we have $1/n\leq 1/N$. Then we have $$\frac{1}{n+1}\leq\frac{1}{n}\leq\frac{1}{N}<\epsilon$$
Then we have $\vert1/(n+1)-0\vert=1/(n+1)<\epsilon$ which shows that $1/(n+1)$ is converged to $0$; thus $b_n$ is converged to $1$.

I am not sure that is valid to do $\{1-\frac{1}{n+1}\}=\{1\}-\{\frac{1}{n+1}\}$ because it looks right to me. If that is not valid, can anyone give me a hit or suggestion to show the sequence is convergent? thanks

Another way is to simply note that, if $n \geq 1$, then
$$
\bigg| \frac{n^{2}}{n^{2}+n} - 1 \bigg| = \frac{n}{n^{2}+n} < \frac{1}{n}.
$$
