第二重要极限
第一重要极限证明: https://www.cnblogs.com/Preparing/p/16548576.html
等比数列求和公式证明： https://www.cnblogs.com/Preparing/p/16581231.html
求证明$$ \lim_{n\to \infty} (1+\frac{1}{n})^{n} 或另一种形式 \lim_{n\to \infty} (1+n)^{\frac{1}{n}} =e $$
思路: 证明数列 \((1+\frac{1}{n})^{n}\) 单调递增且有上界
单调
\[数列x_{n}=(1+\frac{1}{n})^{n}
\\
由二项式定理计算,其数列项T_{1}=T_{2}=1
\]
\[\\ \\
\]
\[T_{3}=\frac{1}{2!}[\frac{n(n-1)}{n^{2}}]=\frac{1}{2!}(1-\frac{1}{n})
\]
\[\\ \\
\]
\[T_{4}=\frac{1}{3!}[\frac{n(n-1)(n-2)}{ n^{3} }]=\frac{1}{3!}[ \frac{(n-1)(n-2)}{n^2} ]=\frac{1}{3!}[\frac{n-1}{n} \cdot \frac{n-2}{n}]=\frac{1}{3!}[(1-\frac{1}{n})(1-\frac{2}{n})]
\]
\[\\ \\
\]
\[T_{5}=\frac{1}{4!}[(1-\frac{1}{n})(1-\frac{2}{n})(1-\frac{3}{n})]
\]
\[\\ \\ ......
\]
\[由上面数列项的规律可推:\
T_{n}=\frac{1}{n!} [(1-\frac{1}{n})(1-\frac{2}{n})(1-\frac{3}{n})...(1- \frac{n-1}{n})]
\]
\[\\ \\
\]
\[x_{n}=T_{1}+T_{2}+T_{3}+...+T_{n}
\]
\[x_{n+1}=(1+\frac{1}{n+1})^{n+1}
\]
\[\\ \\
\]
\[通过二项式定理,得出其数列项H_{1}=H_{2}=1
\]
\[\\ \\
\]
\[H_{3}=C \tbinom {2}{n+1} (\frac{1}{n+1})^{2}=\frac{1}{2!} (\frac{n}{n+1})=\frac{1}{2!} (\frac{n+1-1}{n+1} )=\frac{1}{2!} (1-\frac{1}{n+1})
\]
\[\\ \\
\]
\[H_{4}=C \tbinom {3}{n+1} ( \frac{1}{n+1} )^{3}=\frac{1}{3!} \cdot \frac{n(n-1)}{(n+1)^{2}}
=\frac{1}{3!} (\frac{n}{n+1} \cdot \frac{n-1}{n+1})=\frac{1}{3!}(\frac{n+1-1}{n+1})(\frac{n+1-2}{n+1})
=\frac{1}{3!}(1-\frac{1}{n+1})(1-\frac{2}{n+1})
\]
\[\\ \\
\]
\[同样:
H_{5}=\frac{1}{4!} (1-\frac{1}{n+1})(1-\frac{2}{n+1})(1-\frac{3}{n+1})
\]
\[\\ \\
\]
\[H_{n}=\frac{1}{n!}[(1-\frac{1}{n+1})(1-\frac{2}{n+1})(1-\frac{3}{n+1})...1-\frac{n}{n+1})]
\]
\[\\ \\
\]
\[x_{n+1}=H_{1}+H_{2}+H_{3}+...+H_{n}
\]
\[\\ \\
\]
\[通过比较x_{n}和x_{n+1}发现,除了T_{1} = T_{2} = H_{1} = H_{2}=1之外,H_{3}＞T_{3},H_{4}＞T_{4}, ... , H_{n}＞T_{n}
\]
\[\\ \\
\]
\[\therefore x_{n} < x_{n+1}
\]
\[\\ \\
\]
\[\therefore(1+\frac{1}{n})^{n}是单调递增数列
\]
有界
\[设B=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+...+\frac{1}{n!}
\]
\[\\ \\
\]
\[通过对比每一项的大小,可以发现x_{n}<1+B
\]
\[\\ \\
\]
\[设存在等比数列a_{n}=(\frac{1}{2})^{n-1}(n\in N),公比为\frac{1}{2}
\]
\[\\ \\
\]
\[K=\frac{1}{2^{0}}+\frac{1}{2^{1}}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+...+\frac{1}{2^{n-1}}
\]
\[\\ \\
\]
\[通过比较各项大小进而发现:B<1+K
\]
\[\\ \\
\]
\[\therefore x_{n}<1+K
\]
\[\\ \\
\]
\[K=a_{n}的前n项和,a_{n}首项为1
\]
\[\\ \\
\]
\[由等比数列前n项之和公式,\quad 以及\because x_{n}＜1+K
\]
\[\\ \\
\]
\[\therefore x_{n}< 1+\frac{1 \cdot (1-\frac{1}{2^{n}})}{ 1-\frac{1}{2} }
\\
x_{n}< 1+2- \frac{1}{2^{n-1}}
\]
\[\\
x_{n}< 3- \frac{1}{2^{n-1}}
\\
\therefore 数列x_{n}存在上界
\]
\[\\ \\
\]
\[\because x_{n}=(1+\frac{1}{n})^{n}单调增加且有界\]
\[\\ \\
\]
\[\therefore \lim_{n\to \infty}(1+\frac{1}{n})^{n}=e 证明成立
\]
