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