###### tags: `108上助教` `數學`
# 數學導論HW3
## ch3-2

#### proof
$\mbox{We use the Extended Principle of Mathematical Induction.}$
$\mbox{Let } S =\{n \in \mathbb{N} : n \geq 3 \mbox{ and } (1+\dfrac{1}{n})^n < n\}$
1. $\mbox{Step 1 : check }3 \in S$

$\because (1+\dfrac{1}{3})^3 = \dfrac{64}{27} < 3$
$\therefore 3 \in S$

2. $\mbox{Step 2 : Suppose }n \geq 3 \mbox{ and }n \in S$

$\mbox{Hence we have } (1+\dfrac{1}{n})^n < n\ldots (*)$
$\mbox{We want to show that } (1+\dfrac{1}{n+1})^{n+1} < n+1$ ----(我們的目標)

(proof of this claim)
$\mbox{Note that } \dfrac{1}{n+1} < \dfrac{1}{n} , \mbox{ for } n \in \mathbb{N}$
$\mbox{and if } a < b , a^n < b^n , \mbox{ for } a ,b > 0$
$\mbox{So }$
\begin{equation}
\begin{split}
(1+\dfrac{1}{n+1})^{n+1} &= (1+\dfrac{1}{n+1})^{n}(1+\dfrac{1}{n+1})\\
&< (1+\dfrac{1}{n})^{n} (1+\dfrac{1}{n+1})\\
&< n(1+\dfrac{1}{n+1})\ldots by (*)\\
&= n + \dfrac{n}{n+1}\\
&< n + 1
\end{split}
\end{equation}

$\mbox{Hence } n+1 \in S$
$\mbox{By the Extended Principle of Mathematical Induction, } \{n\in \mathbb{N} : n \geq 3\} \subseteq S\ \Box$