EXERCISE 9.2

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EXERCISE 9.2
1.
\left\{\frac{1}{n}\right\}_{n=1}^{+\infty}

    \[\begin{aligned} & a_{n}=\frac{1}{n}, \quad a_{n+1}=\frac{1}{n+1} \\ \\ & a_{n+1}-a_{n}=\frac{1}{n+1}-\frac{1}{n} \\ \\ & \quad \quad \quad \quad =\frac{n-(n+1)}{n(n+1)} \\ \\ &\quad \quad \quad \quad =\frac{n-n-1}{n(n+1)}=-\frac{1}{n(n+1)}<0 \quad \text { for } n \geqslant 1 \\ \\ & \Rightarrow \quad a_{n+1}-a_{n} < 0 \\ \\ & \Rightarrow \quad a_{n+1}<a_{n} \quad \text { for } n \geqslant 1 \end{aligned}\]

Thus \left\{\frac{1}{n}\right\}_{n=1}^{+\infty} is strictly decreasing.

2.

    \[\begin{aligned} &\left\{1-\frac{1}{n}\right\}_{n=1}^{+\infty} \\ \\ & a_{n}=1-\frac{1}{n}, \quad a_{n+1}=1-\frac{1}{n+1} \\ \\ & a_{n+1}-a_{n}=\left[1-\frac{1}{n+1}\right]-\left[1-\frac{1}{n}\right] \\ \\ &\quad \quad \quad \quad=1-\frac{1}{n+1}-1+\frac{1}{n} \end{aligned}\]

    \[\begin{aligned} & a_{n+1}-a_{n}=\frac{1}{n}-\frac{1}{n+1} \\ \\ &\quad \quad \quad \quad =\frac{n+1-n}{n(n+1)}=\frac{1}{n(n+1)}>0 \quad \text { for } n \geqslant 1 \\ \\ & \Rightarrow \quad a_{n+1}-a_{n}>0 \text { For } n \geqslant 1 \\ \\ & \Rightarrow \quad a_{n+1}>a_{n} \text { For } n \geqslant 1 \end{aligned}\]

Thus \left\{a_{n}\right\} is strictly increasing.

3.

    \[\begin{aligned} & \left\{\frac{n}{2 n+1}\right\}_{n=1}^{+\infty} \\ \\ & a_{n}=\frac{n}{2 n+1} \\ \\ & a_{n+1}=\frac{n+1}{2(n+1)+1}=\frac{n+1}{2 n+2+1} =\frac{n+1}{2 n+3} \end{aligned}\]

    \[\begin{aligned} a_{n+1}-a_{n} & =\frac{n+1}{2 n+3}-\frac{n}{2 n+1} \\ \\ & \quad \quad =\frac{(2 n+1)(n+1)-n(2 n+3)}{(2 n+3)(2 n+1)} \\ \\ & \quad \quad =\frac{2 n^{2}+2 n+n+1-2 n^{2}-3 n}{(2 n+3)(2 n+1)} \end{aligned}\]

    \[\begin{aligned} & a_{n+1}-a_{n}=\frac{1}{(2 n+3)(2 n+1)}>0 \text { For } n \geqslant 1 \\ \\ & \Rightarrow \quad a_{n+1}-a_{n}>0 \text { For } n \geqslant 1 \\ \\ & \Rightarrow \quad a_{n+1}>a_{n} \text { For } n \geqslant 1 \end{aligned}\]

Thus \left\{a_{n}\right\} is strictly increasing.

4.

    \[\begin{aligned} \left\{\frac{n}{4 n-1}\right\}_{n=1}^{+\infty} \\ \\ & a_{n}=\frac{n}{4 n-1} \\ \\ & a_{n+1}=\frac{n+1}{4(n+1)-1}=\frac{n+1}{4 n+4-1} \\ \\ & a_{n+1}=\frac{n+1}{4 n+3} \end{aligned}\]

    \[\begin{aligned} a_{n+1}-a_{n} & =\frac{n+1}{4 n+3}-\frac{n}{4 n-1} \\ \\ &\quad \quad =\frac{(4 n-1)(n+1)-n(4 n+3)}{(4 n+3)(4 n-1)} \\ \\ & \quad \quad =\frac{4 n^{2}+4 n-n-1-4 n^{2}-3 n}{(4 n+3)(4 n-1)} \\ \\ & \quad \quad =\frac{-1}{(4 n+3)(4 n-1)}<0 \text { For } n \geqslant 1 \\ \\ & \Rightarrow \quad a_{n+1}-a_{n}<0 \text { For } n \geqslant 1 \\ \\ & \Rightarrow \quad a_{n+1}<a_{n} \text { For } n \geqslant 1 \end{aligned}\]

Thus \left\{a_{n}\right\} is strictly decreasing

5.

    \[\begin{aligned} &\left\{n-2^{n}\right\}_{n=1}^{+\infty} \\ \\ & a_{n}=n-2^{n}, \quad a_{n+1}=n+1-2^{n+1} \\ \\ & a_{n+1}-a_{n}=n+1-2^{n+1}-n+2^{n} \\ \\ & \quad \quad =1-2^{n+1}+2^{n}=1+2^{n} -2^{n+1}\\ \\ & \quad \quad =1+2^{n}(1-2)=1-2^{n}<0 \text { For } n \geqslant 1 \\ \\ & \Rightarrow \quad a_{n+1}-a_{n}<0 \text { For } n \geqslant 1 \\ \\ & \Rightarrow a_{n+1}<a_{n}$ For $n \geqslant 1 \end{aligned}\]

Thus \left\{a_{n}\right\} is strictly decreasing.

6.

    \[\begin{aligned} & \left\{n-n^{2}\right\}_{n=1}^{+\infty} \\ \\ & a_{n}=n-n^{2}, \quad a_{n+1}=(n+1)-(n+1)^{2} \end{aligned}\]

    \[\begin{aligned} a_{n+1}-a_{n} & =(n+1)-(n+1)^{2}-n+n^{2} \\ \\ & \quad \quad \quad =n+1-\left(n^{2}+2 n+1\right)-n+n^{2} \\ \\ & \quad \quad \quad =1-n^{2}-2 n-1+n^{2} \\ \\ & \quad \quad \quad =-2 n<0 \text { For } n \geqslant 1 \\ \\ & \Rightarrow \quad a_{n+1}-a_{n}<0 \text { for } n \geqslant 1 \\ \\ & \Rightarrow \quad a_{n+1}<a_{n} \text { For } n \geqslant 1 \end{aligned}\]

Thus \left\{a_{n}\right\} is strictly decreasing.

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