###### tags: `108上助教` `數學`
# 數學導論HW2
## ch3-1

#### **proof**
注意這題要證明的是對於所有的正偶數都成立。
我們要用 **the Principle of Mathematical Induction** 來證明。

$\mbox{Note that } n \mbox{ is even if }n = 2k \mbox{ , } k \in \mathbb{N}$

$\mbox{Let }$
$$S = \{k \in \mathbb{N} : (1-\dfrac{1}{2})(1+\dfrac{1}{3})(1-\dfrac{1}{4})\cdots (1-\dfrac{(-1)^{2k}}{2k}) = \dfrac{1}{2}\}$$

$\mbox{We want to show that } S = \mathbb{N}$

$\mbox{First, check } 1 \in S$

$\because (1-\dfrac{1}{2}) = \dfrac{1}{2} \mbox{ , we have } 1 \in S$

$\mbox{Next suppose } k\in S \mbox{ , that is , we have }(1-\dfrac{1}{2})(1+\dfrac{1}{3})(1-\dfrac{1}{4})\cdots (1-\dfrac{(-1)^{2k}}{2k}) = \dfrac{1}{2}$

$\mbox{We need to show that }k+1 \in S$

$$(1-\dfrac{1}{2})(1+\dfrac{1}{3})(1-\dfrac{1}{4})\cdots (1-\dfrac{(-1)^{2k}}{2k})(1-\dfrac{(-1)^{2k+1}}{2k+1})(1-\dfrac{(-1)^{2k+2}}{2k+2})\\
= \dfrac{1}{2} (1-\dfrac{(-1)^{2k+1}}{2k+1})(1-\dfrac{(-1)^{2k+2}}{2k+2})\\
= \dfrac{1}{2} (1-\dfrac{-1}{2k+1})(1-\dfrac{1}{2k+2})\\
=\dfrac{1}{2} (\dfrac{2k+2}{2k+1}) (\dfrac{2k+1}{2k+2})\\
= \dfrac{1}{2}$$

$\mbox{Hence } k+1 \in S$
$\mbox{By the Principle of Mathematical Induction, } S = \mathbb{N}\ \Box$