This is equivalent to show that:

$$\begin{aligned} (\forall \epsilon > 0)(\exists n \in \mathbb{N})(\forall m \ge n)\big[ \space \big\vert \space (\frac{n}{n+1})^2 -1 \space \big\vert < \epsilon \space \big] \end{aligned}$$

Given $\epsilon, 2 > 0$, by the Archimedean property (which says that if $x,y \in \mathbb{R}$ and $x,y > 0$,there is an $n \in \mathbb{N}$ such that $n \cdot x > y$), there is an $n \in \mathbb{N}$ such that:

$$\begin{aligned} (n+1) \cdot \epsilon &> 2 \\ \epsilon &> \frac{2}{n+1} \end{aligned}$$

Consider for all $m \ge n$:

$$\begin{aligned} \big\vert \frac{m^2}{(m+1)^2} - 1 \big\vert = 1 - \frac{m^2}{(m+1)^2} = \frac{2m + 1}{(m+1)^2} \le \frac{2n+1}{(n+1)^2} = \frac{2(n+1) - 1}{(n+1)^2} < \frac{2}{n+1} < \epsilon \end{aligned}$$

$\Box$

# posted by rot13(Unafba Pune) @ 5:01 PM