For \(a_n=\frac{n}{n+1}\), which converges to \(1\), and \(\varepsilon=\frac1{1000}\), what is the smallest \(N\) such that \(|a_n-1|<\varepsilon\) for all \(n\ge N\)?
\(|a_n-1|=\frac1{n+1}<\frac1{1000}\) is equivalent to \(n+1>1000\), that is \(n\ge1000\). For \(n=999\) we get exactly \(\frac1{1000}\), which is not smaller.
A sequence \((a_n)\) converges to a number \(L\) if, for every tolerance \(\varepsilon>0\), there is an index \(N\) from which all terms are within distance \(\varepsilon\) of \(L\).