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For each positive integer \(n\), let \(a_n=n/(n+2)\). Prove from the definition of convergence that \(a_n\to1\).
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The definition asks: given any \(\varepsilon>0\), can you make \(|a_n-1|<\varepsilon\) for every sufficiently large \(n\)? First simplify \(|a_n-1|\).
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Let \(\varepsilon>0\). For every positive integer \(n\), \[ \left|\frac{n}{n+2}-1\right|=\left|\frac{-2}{n+2}\right|=\frac{2}{n+2}. \] Choose a positive integer \(N>2/\varepsilon\). Such an integer exists because the positive integers are unbounded. If \(n\ge N\), all denominators are positive, so \[ \frac{2}{n+2}\le\frac{2}{N+2}<\frac{2}{N}<\varepsilon. \] Thus, for every positive tolerance \(\varepsilon\), this \(N\), chosen for the given \(\varepsilon\), works for every \(n\ge N\). This is exactly the definition of \(a_n\to1\).