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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).