Limit of a sequence - Example

Te-Sheng Lin (林得勝)

Last updated on Mar 6, 2020 1 min read calculus

Section 11.1, exercise 79: Find the limit of ${\sqrt{2}, \sqrt{2 \sqrt{2}}, \sqrt{2 \sqrt{2 \sqrt{2}}}, \dots}$

Remark: 這個數列也可以用以下表示方式 $a_{1} = \sqrt{2}, a_{n + 1} = \sqrt{2 a_{n}}, n \in N, n > 1.$

Method 1

我們首先證明它有極限
- $a_{n} < 2$ (我們使用數學歸納法)
  1. 已知 $a_{1} = \sqrt{2} < 2$ , 成立.
  2. 假設 $a_{k} < 2$ , 則 $a_{k + 1} = \sqrt{2 a_{k}} < \sqrt{2 \times 2} = 2$ , 亦成立.
  3. 根據數學歸納法我們得到 $a_{n} < 2$ , $\forall n$ .
- $a_{n} > 0$
  
  這個顯然
- claim: $a_{n + 1} > a_{n}$
  
  由於 $0 < a_{n} < 2$ , 我們可以得到 $a_{n} (a_{n} - 2) < 0$ , 移項並兩邊同時開根號得到 $a_{n + 1} = \sqrt{2 a_{n}} > a_{n}$ .
- 因為數列 ${a_{n}}$ 遞增且有界, 根據 Monotone sequence theorem 此數列收斂.
我們接著求此數列極限值

假設 $lim_{n \to \infty} a_{n} = L$ , 則 $lim_{n \to \infty} a_{n + 1} = lim_{n \to \infty} \sqrt{2 a_{n}}, ⟹ L = \sqrt{2 L} .$ 所以 $L = 0$ 或 $L = 2$ . 我們知道 $a_{n}$ 為遞增數列所以 $L = 2$ .

Method 2

這是看 slader 學會的解法.

$a_{1} = \sqrt{2} = 2^{1 / 2}$
$a_{2} = \sqrt{2 a_{1}} = 2^{1 / 2 + 1 / 4}$
$a_{3} = \sqrt{2 a_{2}} = 2^{1 / 2 + 1 / 4 + 1 / 8}$

觀察發現此數列收斂到 $2^{\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots} = 2^{1} = 2.$

sequence

Te-Sheng Lin (林得勝)

Professor

The focus of my research is concerned with the development of analytical and computational tools, and further to communicate with scientists from other disciplines to solve engineering problems in practice.