SciTech-Mathmatics-Analysis-Infinite Series+Limit: 无穷级数+极限: $\large e = \lim{(1+\frac{1}{n})^n}$

SciTech-Mathmatics-Analysis

• Infinite Series: 无穷级数
• Limit: 极限：
  \(\large e = \underset{n \rightarrow \infty}{\lim} {(1+\frac{1}{n})^n}\)
  这是一个非常有现实意义的无穷级数：
  假设：
  一家银行设计，存一年利息率是100%; 这"一年"允许分成任意期数\(\large n\)；
  求\(\large n\)取多少期时，总利息收益最大？
  \(\large \begin{array} {rll} \\ 一年分 1期 & Total &= (1+ \frac{1}{1})^1 &= 2.0 \\ 一年分 2期 & Total &= (1+ \frac{1}{2})^2 &= 2.25 \\ 一年分 3期 & Total &= (1+ \frac{1}{3})^3 &= 2.37 \\ \cdots & & & \\ 一年分 n期 & Total &= (1+ \frac{1}{n})^n &= e = 2.718 \cdots \\ \end{array}\)

求导数: \(\large \underset{x \rightarrow \infty}{\lim} {(1+\frac{1}{x})^x}\)

\(\large \begin{array} {rrl} \\ \because & f(x) &= (1+ \frac{1}{x})^x ,\ where \ x \in [1, +\infty] \\ & &= e^{x \cdot {ln(\frac{x+1}{x})} } \\ \therefore & f'(x) &= e^{x \cdot {ln(\frac{x+1}{x})} } \cdot \{[\ 1 \cdot ln(\frac{x+1}{x})\ ] + x \cdot [\ \frac{x}{x+1} \cdot ( - x^{-2} ) \ ]\} \\ & &=(1+ \frac{1}{x})^x \cdot (\ ln(1+\frac{1}{x}) - \frac{1}{x+1} \ ) \\ let & f'(x) &= 0 \\ & \rightarrow & ln(1+ \frac{1}{x}) = \frac{1}{x+1} \\ & \rightarrow & \frac{x+1}{x} = e^{\frac{1}{x+1}} \\ & \rightarrow & \frac{x}{x+1} = 1 - \frac{1}{x+1} = e^{(- \frac{1}{x+1})} = (e^{\frac{1}{x+1}})^{-1} \\ & let & y = - \frac{1}{x+1} \in [-\frac{1}{2}, 0) ,\text{ since } \ x \in [1, +\infty] \\ & \rightarrow & y+1 = e^y , \text{ where } \ y \in [-\frac{1}{2}, 0) \\ & & \text{ there's only one cross point } (0, 1) \text{ of } z=(y+1) \text{ and } z= e^y \\ & & \text{ since } \frac{d(y+1)}{dy} =1 > e^{y} = \frac{d(e^y)}{dy} \text{ when } \ y \in [-\infty, 0) \\ & \rightarrow & y+1 < e^y , \text{ when } \ y \in [-\frac{1}{2}, 0) \\ & \rightarrow & \frac{x}{x+1} < (e^{\frac{1}{x+1}})^{-1} , \text{ when } \ x \in [1, +\infty] \\ & \rightarrow & \frac{x+1}{x} > e^{\frac{1}{x+1}}, \text{ when } \ x \in [1, +\infty] \\ & \rightarrow & ln(1+ \frac{1}{x}) > \frac{1}{x+1}, \text{ when } \ x \in [1, +\infty] \\ & \rightarrow & f'(x) > 0, \text{ when } \ x \in [1, +\infty] \\ \therefore & f(x) &= (1+ \frac{1}{x})^x \ 单调递增,\ where \ x \in [1, +\infty] \\ \end{array}\)

求极限: \(\large \underset{x \rightarrow \infty}{\lim} {(1+\frac{1}{x})^x}\)

\(\large \underset{n \rightarrow \infty}{\lim} {(1+\frac{1}{x})^x} = \underset{x \rightarrow \infty}{\lim} {\frac{(x+1)^x}{x^x}} = \underset{x \rightarrow \infty}{\lim} {\frac{ e^{x \cdot ln(x+1)} }{ e^{x \cdot ln(x)}} } = \underset{x \rightarrow \infty}{\lim} {e^{x \cdot ln(x+1) - x \cdot ln(x)} }\)

转化为求极限：\(\large \underset{x \rightarrow \infty}{\lim} {x \cdot ( ln(x+1) - ln(x) ) }\)

也就是求极限(\(\large \infty \cdot 0\ 化为\ \frac{0}{0}\)型)：\(\large \underset{x \rightarrow \infty}{\lim} {\frac{( ln(x+1) - ln(x) )}{\frac{1}{x}} }\)

\(\large \begin{array} {rrl} \\ let & g(x) &= ln(x+1) - ln(x) > 0 \\ & \rightarrow & g(1) = ln 2 \in (0, 1) \\ & & g(e-1) = 1-ln(e-1) \in (0, 1) \\ & & g(e) = ln(e+1) - 1 \in (0, 1) \\ & g'(x) &= \bm{ \frac{-1}{x(x+1)} } < 0 , \ when\ x \in (0,+\infty) \\ & \rightarrow & g(x) 单调递减 \ and \ g(x) > 0, \ when\ x \in (0,+\infty) \\ \therefore & g(x) & 极限存在; \\ & \exists\ \epsilon &= \frac{1}{ e^{\sigma} - 1}, \ when \ x > \epsilon \\ & \forall\ \sigma &> 0,\ | g(x) - 0 | < \sigma \\ & |g(x)| &=[ ln(x+1) - ln(x) ] < \sigma \\ & \uparrow & e^ {[ln(x+1)- ln(x)]} < e^{\sigma} \\ & & 1 + \frac{1}{x} < e^{\sigma} \\ & & x > \frac{1}{ e^{\sigma} - 1} \\ \therefore & \underset{x \rightarrow \infty}{\lim} {g(x)} &= \underset{x \rightarrow \infty}{\lim} {( ln(x+1) - ln(x) ) } = 0 \\ \end{array}\)

\(\large \begin{array} {rrl} \\ now & \text{ we can use } & \text{ the } L'Hospital\ Law \text{ for resolving } \frac{0}{0} \text{ form limit } \\ \end{array}\)

\(\large \begin{array} {ll} \\ \underset{x \rightarrow \infty}{\lim} {\frac{( ln(x+1) - ln(x) )}{\frac{1}{x}} } &=\underset{x \rightarrow \infty}{\lim} { \frac{( \frac{1}{x+1} - \frac{1}{x} )}{\frac{-1}{x^2}} }, \ L'Hospital\ Law \\ &=\underset{x \rightarrow \infty}{\lim} { \frac{\frac{-1}{x(x+1)}}{\frac{-1}{x^2}} } =\underset{x \rightarrow \infty}{\lim} { \frac{x}{x+1} } =\underset{x \rightarrow \infty}{\lim} { (1 - \frac{1}{x+1} )} \\ &= 1 \\ \end{array}\)

\(\large \begin{array} {ll} \\ Sum \ up: & \\ \underset{n \rightarrow \infty}{\lim} {(1+\frac{1}{x})^x} & = \underset{x \rightarrow \infty}{\lim} {e^{x \cdot ln(x+1) - x \cdot ln(x)} } = e^{\underset{x \rightarrow \infty}{\lim}{x \cdot ( ln(x+1) - ln(x) )} } \\ &= e^{1} = e \\ Finally:& \\ \underset{n \rightarrow \infty}{\lim} {(1+\frac{1}{x})^x} = e \\ \\ \end{array}\)