Show that:

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 $0$.