EXERCISE 9.1

by

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.

Leave a Comment