最初的定义及计算方法
e = lim n → ∞ ( 1 + 1 n ) n = lim n → ∞ ∑ i = 0 n C i n 1 n − i ( 1 n ) i = lim n → ∞ [ C 0 n 1 n ( 1 n ) 0 + C 1 n 1 n − 1 ( 1 n ) 1 + C 2 n 1 n − 2 ( 1 n ) 2 + C 3 n 1 n − 3 ( 1 n ) 3 + … + C n n 1 0 ( 1 n ) n ] = lim n → ∞ [ 1 × 1 + n × 1 n + n ! ( n − 2 ) ! 2 ! × 1 n 2 + n ! ( n − 3 ) ! 3 ! × 1 n 3 + … + 1 × 1 n n ] = lim n → ∞ [ 1 + 1 + n × ( n − 1 ) 2 n 2 + n × ( n − 1 ) ( n − 2 ) 3 × 2 n 3 + … + 1 n n ] = 2 + 1 2 + 1 6 + … = 2.71828 ⋯ \begin{aligned} e&=\lim_{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{n} \\ &=\lim_{n \rightarrow \infty} \sum_{i=0}^{n} C_{i}^{n} 1^{n-i}\left(\frac{1}{n}\right)^{i} \\ &=\lim_{n \rightarrow \infty}\left[C_{0}^{n} 1^{n}\left(\frac{1}{n}\right)^{0}+C_{1}^{n} 1^{n-1}\left(\frac{1}{n}\right)^{1}+C_{2}^{n} 1^{n-2}\left(\frac{1}{n}\right)^{2}+C_{3}^{n} 1^{n-3}\left(\frac{1}{n}\right)^{3}+\ldots+C_{n}^{n} 1^{0}\left(\frac{1}{n}\right)^{n}\right] \\ &=\lim_{n \rightarrow \infty}\left[1 \times 1+n \times \frac{1}{n}+\frac{n !}{(n-2) ! 2 !} \times \frac{1}{n^{2}}+\frac{n !}{(n-3) ! 3 !} \times \frac{1}{n^{3}}+\ldots+1 \times \frac{1}{n^{n}}\right] \\ &=\lim_{n \rightarrow \infty}\left[1+1+\frac{n \times(n-1)}{2 n^{2}}+\frac{n \times(n-1)(n-2)}{3 \times 2 n^{3}}+\ldots+\frac{1}{n^{n}}\right] \\ &=2+\frac{1}{2}+\frac{1}{6}+\ldots \\ &=2.71828 \cdots \end{aligned}\\ e=n→∞lim(1+n1)n=n→∞limi=0∑nCin1n−i(n1)i=n→∞lim[C0n1n(n1)0+C1n1n−1(n1)1+C2n1n−2(n1)2+C3n1n−3(n1)3+…+Cnn10(n1)n]=n→∞lim[1×1+n×n1+(n−2)!2!n!×n21+(n−3)!3!n!×n31+…+1×nn1]=n→∞lim[1+1+2n2n×(n−1)+3×2n3n×(n−1)(n−2)+…+nn1]=2+21+61+…=2.71828⋯
补点能算出 e e e的等式
极限相关
lim n → ∞ ( 1 + 1 n ) n = lim t → 0 ( 1 + t ) 1 t = e lim n → ∞ n ! n n = e − 1 lim n → ∞ ( ( ( n + 1 ) ! ) 1 n + 1 − ( n ! ) 1 n ) = 1 e lim n → ∞ ( ∏ i = 1 n p i ) 1 p n = e \begin{aligned} \lim_{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{n} =\lim_{t \rightarrow 0}(1+t)^{\frac{1}{t}}&=e\\ \lim\limits_{n\to\infty}\frac{\sqrt[n]{n!}}{n}&=e^{-1}\\ \lim\limits_{n\rightarrow\infty} \left( ((n+1)!)^{\frac{1}{n+1}}-(n!)^{\frac{1}{n}} \right)&=\frac{1}{e} \end{aligned}\\\lim _{n \rightarrow \infty}\left(\prod_{i=1}^{n} p_{i}\right)^{\frac{1}{p_{n}}}=e \\ n→∞lim(1+n1)n=t→0lim(1+t)t1n→∞limnnn!n→∞lim(((n+1)!)n+11−(n!)n1)=e=e−1=e1n→∞lim(i=1∏npi)pn1=e
p n p_{n} pn是第 n n n个素数
补几个简单拓展
lim n → + ∞ ( n n ⋅ ( n + 1 ) ! n + 1 − ( n + 1 ) n + 1 ⋅ n ! n ) = 1 e lim n → ∞ ( n 1 + n − 1 2 + ⋯ + 1 n ln ( n ! ) ) ln ( n ! ) n = e γ lim n → ∞ ( ( n + 1 ) ! n + 1 n ! n ) n ＝ e \begin{aligned} \lim_{n\to+\infty}\left(\sqrt[n]n\cdot\sqrt[n+1]{(n+1)!}-\sqrt[n+1]{(n+1)}\cdot\sqrt[n]{n!}\right)&=\frac{1}{e}\\ \lim_{n\rightarrow\infty}\left(\frac{\frac{n}{1}+\frac{n-1}{2}+\cdots+\frac{1}{n}}{\ln(n!)} \right)^{
{\frac{\ln(n!)}{n}}}&=e^\gamma \\ \lim_{n\to\infty}\left(\frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}}\right)^n&＝e\\ \end{aligned}\\ n→+∞lim(nn⋅n+1(n+1)!−n+1(n+1)⋅nn!)n→∞lim(ln(n!)1n+2n−1+⋯+n1)nln(n!)n→∞lim(nn!n+1(n+1)!)n=e1=eγ＝e
级数相关
连分数
e = 2 + 1 1 + 1 2 + 1 1 + 1 1 + 1 4 + 1 1 + 1 1 + 1 6 + 1 1 + ⋱ e=2+\frac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{1+\frac{1}{4+\frac{1}{1+\frac{1}{1+\frac{1}{6+\frac{1}{1+\ddots}}}}}}}}} \\ e=2+1+2+1+1+4+1+1+6+1+⋱111111111
两个小趣闻
( i ) \quad(\mathrm{i}) (i)在 G o o g l e Google Google 2004 2004 2004年的首次公开募股，集资额不是通常的整数，而是 2718281828 2718281828 2718281828美元
( i i ) \quad(\mathrm{ii}) (ii)著名计算机科学家高德纳的软件 M e t a f o n t Metafont Metafont版本号趋向 e e \,\, e(就是说版本号是 2 2 2， 2.7 2.7 2.7， 2.71 2.71 2.71， 2.718 2.718 2.718， ⋯ \cdots ⋯)，与之相对的是：TeX的版本号是趋向于圆周率 π \pi π
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