1.7 Answer

1.7

1.

(a)For any number $N$ is chosen, there is a number $n$ so that $2^{n} \geq N$. Thus, it is unbounded.

(b) $a_{n}$ represents $\sqrt{2}$ to the $n$th decimal place. So even as $n$ gets larger, it can not be greater than $\sqrt{2}$. Thus it is bounded.

2.

(a)Let $\lim_{n \to \infty}a_{n} = \alpha$. Then $a_{n+1} = \sqrt{3a_{n} + 4}$. Thus $\alpha = \sqrt{3\alpha + 4}$． Now square both sides, we have $\alpha^2 = 3 \alpha + 4$. Thus, $\alpha = -1,4$． By the initial condition $a_{1} = 1$, then $\alpha = 4$．We next show that $\lim_{n \to \infty}a_{n} = 4$． By the theorem1．13, we need to show that the exists $0 < \lambda < 1$ so that $\vert a_{n+1} - 4\vert \leq \lambda \vert a_{n} - 4\vert$.

$$\vert a_{n+1} - 4\vert = \vert\sqrt{3a_{n} + 4} - 4\vert = \vert\frac{3a_{n} + 4 - 4^2}{\sqrt{3a_{n} + 4} + 4}\vert$$

$$= \vert\frac{3(a_{n} - 4)}{\sqrt{3a_{n} + 4} + 4}\vert = \frac{3}{\sqrt{3a_{n} + 4} + 4} \vert a_{n} - 4\vert$$

ここで， $\lambda = \frac{3}{\sqrt{3a_{n} + 4} + 4} < 1$

(b) $${a_{3} = 2^{\frac{1}{2}}, a_{4} = 2^{\frac{3}{4}}, a_{5} = 2^{\frac{5}{8}}, \ldots , a_{n} = 2^{\frac{2(n-2) - 1}{2^{n-2}}}}$$． Thus, $\lim_{n \to \infty} a_{n} = 1$

3.

(a) $${\lim_{n \to \infty}(1 - \frac{1}{n^2})^n = \lim_{n \to \infty}(1 + \frac{1}{n})^{n}(1 - \frac{1}{n})^{n} = e \frac{1}{e} = 1}$$

(b)Let $\frac{2}{n} = \frac{1}{m}$. Then $n = 2m$. Thus,

$$\lim_{n \to \infty}(1 + \frac{2}{n})^{n} = \lim_{m \to \infty}[(1 + \frac{1}{m})^{m}]^{2} = e^{2}$$

(c)Using $${\lim_{n \to \infty}\left\vert\frac{a_{n+1}}{a_{n}}\right\vert < 1 \leftrightarrow \lim_{n \to \infty} a_{n} = 0 }$$, we have

$$\lim_{n \to \infty} \left\vert\frac{a_{n+1}}{a_{n}}\right\vert = ... ... \infty}\frac{2^{n+1}/(n+1)!}{2^{n}/n!} = \lim_{n \to \infty}\frac{2}{n+1} = 0$$

よって $${\lim_{n \to \infty} \frac{2^{n}}{n!} = 0}$$

(d) Using $${\lim_{n \to \infty}\left\vert\frac{a_{n+1}}{a_{n}}\right\vert < 1 \leftrightarrow \lim_{n \to \infty} a_{n} = 0 }$$, we have

$$\lim_{n \to \infty} \left\vert\frac{a_{n+1}}{a_{n}}\right\vert = \lim_{n \to \infty}\frac{(n+1)!/(n+1)^{n+1}}{n!/n^{n}}$$

$$= \lim_{n \to \infty}\frac{n+1}{n+1}\left(\frac{n}{n+1}\right)^{n}$$

$$= \lim_{n \to \infty}\left(\frac{1}{1+\frac{1}{n}}\right)^{n} = \frac{1}{e}$$

よって $${\lim_{n \to \infty} \frac{n!}{n^{n}} = 0}$$