The sequence \(\{ s_{n}\} =\left\{ \dfrac{3n}{n+2}\right\} \) is bounded both from above and below because \[ \dfrac{3n}{n+2}=\dfrac{3}{1+\dfrac{2}{n}}<3 \qquad \hbox{ and }\qquad \dfrac{3n}{n+2}>0 \quad \hbox{ for all } n\geq 1 \] See Figure 12(a).

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The sequence \(\{a_{n}\} =\left\{ \dfrac{4n}{3}\right\}\) is bounded from below because \(\dfrac{4n}{3}>1\) for all \(n\geq 1.\) It is not bounded from above because \(\lim\limits_{n\rightarrow \infty }\dfrac{4n}{3}=\dfrac{4}{3}\lim\limits_{n\rightarrow \infty }n=\infty.\) See Figure 12(b).

The sequence \(\{b_{n}\} =\left\{ (-1)^{n+1}n\right\}\) is neither bounded from above nor bounded from below.If \(n\) is odd, \(\lim\limits_{n\rightarrow \infty }b_{n} = \lim\limits_{n\rightarrow \infty } n =\infty\), and if \(n\) is even, \(\lim\limits_{n\rightarrow \infty }b_{n}= \lim\limits_{n\rightarrow \infty } (-n) = -\infty.\) See Figure 12(c).