EXERCISE 9.1

EXERCISE 9.1

1.

(a)

$$
\begin{aligned}
& 1, \frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots \\ \\
& a_{1}=1, \quad a_{2}=\frac{1}{3}, \quad a_{3}=\frac{1}{3^{2}}, a_{4}=\frac{1}{3^{3}}, \ldots \\ \\
& a_{1}=1, \quad a_{2}=\frac{1}{3^{2-1}}, \quad a_{3}=\frac{1}{3^{3-1}}, a_{4}=\frac{1}{3^{4-1}}, \ldots \\ \\
& \ldots \quad a_{n}=\frac{1}{3^{n-1}}, \ldots
\end{aligned}
$$

(b)

$$
\begin{aligned}
& 1, \quad-\frac{1}{3}, \frac{1}{9},-\frac{1}{27}, \cdots \\ \\
& a_{1}=(-1)^{1-1}, \quad a_{2}=(-1)^{2-1} \frac{1}{3^{2-1}}, a_{3}=(-1)^{3-1} \frac{1}{3^{3-1}} \\ \\
& a_{4}=(-1)^{4-1} \frac{1}{3^{4-1}}, \cdots a_{n}=(-1)^{n-1} \frac{1}{3^{n-1}}
\end{aligned}
$$

(c)

$$
\begin{aligned}
& \frac{1}{2}, \frac{3}{4}, \frac{5}{6}, \frac{7}{8}, \cdots \\ \\
& a_{1}=\frac{1}{2}, a_{2}=\frac{3}{4}, a_{3}=\frac{5}{6}, a_{4}=\frac{7}{8}, \\ \\
& , a_{n}=\frac{2 n-1}{2 n}, \cdots
\end{aligned}
$$

(d)

$$
\begin{aligned}
& \frac{1}{\sqrt{\pi}}, \frac{4}{\sqrt[3]{\pi}}, \frac{a}{\sqrt[4]{\pi}}, \frac{16}{\sqrt[5]{\pi}}, \ldots \\ \\
& a_{1}=\frac{1}{\sqrt{\pi}}, a_{2}=\frac{4}{\sqrt[3]{\pi}}, a_{3}=\frac{9}{4 \sqrt{\pi}}, a_{n}=\frac{16}{\sqrt[5]{\pi}}, \ldots \\ \\
& a_{1}=\frac{(1)^{2}}{(\pi)^{\frac{1}{2}}}, a_{2}=\frac{2^{2}}{(\pi)^{\frac{1}{2+1}}}, a_{3}=\frac{3^{2}}{(\pi)^{\frac{1}{3+1}}}, a_{4}=\frac{4^{2}}{(\pi)^{\frac{1}{4+1}}}, \ldots \\ \\
& a_{n}=\frac{n^{2}}{(\pi)^{\frac{1}{n+1}}}, \ldots
\end{aligned}
$$

2.

(a) $$ \begin{aligned}

&1,-r, r^{2},-r^{3}, \ldots \end{aligned} $$

$$
\begin{aligned}
& a_{1}=1, \quad a_{2}=-r, \quad a_{3}=r^{2}, \quad a_{4}=-r^{3}, \ldots \\ \\
& a_{1}=(-1)^{1-1} 1, \quad a_{2}=(-1)^{2-1} r^{2-1}, \quad a_{3}=(-1)^{3-1} r^{3-1}, \quad a_{4}=(-1)^{4-1} r^{4-1}, \ldots \\ \\
& \ldots a_{n}=(-1)^{n-1} r^{n-1} \ldots \\
\end{aligned}
$$

Then $a_{n}=(-1)^{n-1} r^{n-1}$ \quad for $n=1,2,3, \Idots$.

And $$a_{1}=1, a_{2}=-r, a_{3}=r^{2}, \quad a_{4}=-r^{3}, \ldots$$

$$
\begin{aligned}

& a_{1}=(-1)^{0} 1^{0}, \quad a_{2}=(-1)^{1} r^{1}, \quad a_{3}=(-1)^{2} r^{2}, \quad a_{4}=(-1)^{3} r^{3}, \Idots \\ \\
& \text {Thus} \quad a_{n}=(-1)^{n} r^{n}=(-r)^{n} \quad \text { for } n=0,1,2, \ldots
\end{aligned}
$$

(b)

$$
\begin{aligned}
& r,-r^{2}, r^{3},-r^{4}, \ldots \\ \\
& a_{1}=r, a_{2}=-r^{2}, a_{3}=r^{3}, a_{4}=-r^{4}, \ldots \\ \\

& a_{1}=-(-r), a_{2}=-(-r)^{2}, a_{3}=-(-r)^{3}, a_{4}=-(-r)^{4}, \Idots\\ \\
& \text { Thus } \quad a_{n}=-(-r)^{n}, \quad \text { for } n=1,2,3 \ldots
\end{aligned}
$$

and

$$
\begin{aligned}

& a_{1}=r, \quad a_{2}=-r^{2}, \quad a_{3}=r^{3}, a_{4}=-r^{4}, \ldots \\ \\
& a_{1}=(-1)^{0} r^{0+1}, \quad a_{2}=(-1)^{1} r^{1+1}, \quad a_{3}=(-1)^{2} r^{2+1}, a_{4}=(-1)^{3} r^{3+1},
& \ldots \\ \\
& \text {Thus} \quad a_{n}=(-1)^{n+1} r^{n+1} \text { for } n=0,1,2,3 \ldots
\end{aligned}
$$

3.

(a)

$$
\begin{aligned}
& \left\{1+(-1)^{n}\right\} \quad n=0,1,2,3, \Idots \\ \\
& a_{n}=1+(-1)^{n} \quad n=0,1,2,3, \Idots \\ \\
& a_{0}=1+(-1)^{0}=1+1=2 \\ \\
& a_{1}=1+(-1)^{1}=1-1=0 \\ \\
& a_{2}=1+(-1)^{2}=1+1=2 \\ \\
& a_{3}=1+(-1)^{3}=1-1=0
\end{aligned}
$$

(b)

$$
\begin{aligned}
& \{\cos n \pi\} \quad n=0,1,2,3, \ldots \\ \\
& a_{n}=\cos n \pi, n=0,1,2,3, \ldots \\ \\
& a_{0}=\cos 0=1 \\ \\
& a_{1}=\cos \pi=-1 \\ \\
& a_{2}=\cos 2 \pi=1 \\ \\
& a_{3}=\cos 3 \pi=-1
\end{aligned}
$$

(c)

$$
4,0,4,0, \ldots
$$

$$
a_{n}=2+(-2)^{n}, \quad n=0,1,2, \ldots
$$

and

$$
a_{n}=2+2 \cos n \pi, \quad n=0,1,2,3, \ldots
$$

4.

$$
\begin{aligned}
& 1 \cdot 2, \quad 1 \cdot 2 \cdot 3 \cdot 4, \quad 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6,1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8, \ldots \\ \\
& 2 !, \quad 4 !, \quad 6 !, \quad 8 !, \ldots \\ \\
& \text {Thus general term is} \quad (2n)!
\end{aligned}
$$

(b)

$$
\begin{aligned}
& 1,1 \cdot 2 \cdot 3,1 \cdot 2 \cdot 3 \cdot 4 \cdot 5,1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7, \ldots \\ \\
& 1 !, \quad 3 !, \quad 5 !, \quad 7 !, \ldots \\ \\
& \text {Thus general term is} (2n-1)!
\end{aligned}
$$

5.

$$
f(x)=\cos \left(\frac{\pi}{2} x\right)
$$

(a) $$\quad f(n)=\cos \left(\frac{\pi}{2} n\right)$$

For $n=1,2,3, \ldots$ we have

$$
\begin{aligned}
& f(1)=\cos \frac{\pi}{2}=0 \\ \\
& f(2)=\cos \left(\frac{\pi}{2} \cdot 2\right)=\cos \pi=-1 \\ \\
& f(3)=\cos \left(\frac{\pi}{2} 3\right)=0 \\ \\
& f(4)=\cos \left(\frac{\pi}{2} \cdot 4\right)=\cos (2 \pi)=1 \\ \\
& f(5)=\cos \left(\frac{\pi}{2} 5\right)=\cos \left(5 \frac{\pi}{2}\right)=0 \\ \\
& f(6)=\cos \left(\frac{\pi}{2} 6\right)=\cos (3 \pi)=-1
\end{aligned}
$$

We see that the $f(n)$ oscillates between $0$ and $\pm 1$ , therefore $\operatorname{lim}_{x \rightarrow+\infty} f(x)$ does not exist.

(b)

$$
\begin{aligned}
& a_{n}=f(2 n) \\ \\
& a_{n}=\cos \left(\frac{\pi}{2} 2 n\right)=\cos (n \pi) \quad n=1,2,3, \cdots \\ \\
& a_{1}=\cos (\pi)=-1, \quad a_{2}=\cos (2 \pi)=1 \\ \\
& a_{3}=\cos (3 \pi)=-1, \quad a_{4}=\cos (4 \pi)=1 \\ \\
& a_{5}=\cos (5 \pi)=-1
\end{aligned}
$$

(C)

$$
\begin{aligned}
& a_{n}=\cos (n \pi) \\ \\
& \lim _{n \rightarrow+\infty} a_{n}=\lim _{n \rightarrow+\infty}(\cos n \pi) .
\end{aligned}
$$

Since $a_{n}$ oscillates between +1 and -1 , therefore it does not converge.

6.
(a)

$$
\begin{aligned}
& b_{n}=f(2 n+1)=\cos \left(\frac{\pi}{2}(2 n+1)\right) \\ \\
& b_{n}=\cos (2 n+1) \frac{\pi}{2}, n=1,2,3, \cdots \\ \\
& b_{1}=\cos \left(3 \frac{\pi}{2}\right)=0 \\ \\
& b_{2}=\cos \left(5 \frac{\pi}{2}\right)=0 \\ \\
& b_{3}=\cos \left(7 \frac{\pi}{2}\right)=0 \\ \\
& b_{4}=\cos \left(9 \frac{\pi}{2}\right)=0
\end{aligned}
$$

(b) We see from part (a) that

$$
b_{n}=0 \quad \text { for } n=1,2,3, \cdots
$$

Thus $\left\{b_{n}\right\}$ converges to 0 .

(c)

$$
\begin{array}{ll}
f(n)=\cos \left(n \frac{\pi}{2}\right) \quad n=1,2,3, \cdots \\ \\
f(1)=\cos \left(\frac{\pi}{2}\right)=0, \quad f(2)=\cos (\pi)=-1 \\ \\
f(3)=\cos \left(3 \frac{\pi}{2}\right)=0, \quad f(4)=\cos (2 \pi)=1, \ldots .
\end{array}
$$

We observe that $f(n)$ oscillates between $\pm 1$ and $0$ , therefore $f(n)$ does not converge.

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