Showing a Sequence Converges

Show that:

1. \(\lim\limits_{n\rightarrow \infty }c=c\)
2. \(\lim\limits_{n\rightarrow \infty }\dfrac{1}{n}=0\)

Solution (a) The graph of the sequence \(\{ s_{n}\} =\{c\}\) suggests that \(\{ s_{n}\} \) converges to \(c.\) See Figure 4.

To show the sequence converges to \(c,\) we look at \(\vert s_{n}-c\vert .\) Then for any \(\varepsilon >0,\) \[ \vert s_{n}-c\vert =\left\vert c-c\right\vert =0<\varepsilon \qquad \hbox{ for all }n \]

The sequence \(\{ s_{n}\} =\{c\} \) converges to \(c.\)

(b) The graph of the sequence \(\{s_{n}\} =\left\{\dfrac{1}{n}\right\} \) shown in Figure 5 suggests that \(\{ s_{n}\} \) converges to \(0.\)

To show the sequence \(\{ s_{n}\} =\left\{ \dfrac{1}{n}\right\}\) converges to \(0,\) we look at \[ \left\vert s_{n}-0\right\vert =\left\vert \dfrac{1}{n}-0\right\vert =\dfrac{1}{n} \]

For any \(\varepsilon > 0\), choose any integer \(N>\dfrac{1}{\varepsilon }\). Then for all \(n>N>\dfrac{1}{\varepsilon },\) we have \(\left\vert s_{n}-0\right\vert =\dfrac{1}{n}<\dfrac{1}{N}<\varepsilon\), so the sequence \(\{ s_{n}\} =\left\{ \dfrac{1}{n}\right\}\) converges to \(0\).