(quadquad前言quadquad\)
(此证明,改编自中科大数分教材,史济怀版\)
(中科大教材,用的是先固定m,再放大m,跟菲赫金哥尔茨的方法一样。\)
(而我这里的证明,是依据m的任意性,后来发现小平邦彦的《微积分入门》里,也是用的这个方法,即,m的任意性。\)
(中科大和菲赫金哥尔茨用的记号是a_{m},我在知乎咨询龚漫奇老师后,根据龚老师的建议,改为a_{n,m},以避免\)
(混淆,否则a_{m},相当于a_{n}的n取值m,只有一个变量n,取值m,而a_{n}{m}有两个变量m,n\)
(对e_{n,m}取极限时,相当于二元二次极限(注意,非二重极限),即n,m,一先一后取极限,而非二重极限\)
(同时,我在证明中明确了数列极限的保不等式性的应用,\)
(用了两次数列保不等式性,把e当做常数数列。\)
(中科大和菲赫金哥尔茨的先固定m,对n取极限之后,再对m取极限,本质上就是二元二次极限,但是并未明确提及\)
(二元二次极限这个概念)
(------------------------------------------------------------\)
(记S_{n}=frac{1}{0!}+frac{1}{1!}+frac{1}{2!}+cdotcdotcdot+frac{1}{n!})
(显然,S_{n}是递增数列, 且)
(S_{n}leqslant1+1+frac{1}{2}+frac{1}{2^2}+cdotcdotcdot+frac{1}{2^(n-1)}<3)
(显然,S_{n}是递增)
(因为当n趋于无穷时,1+1+frac{1}{2}+frac{1}{2^2}+cdotcdotcdot+frac{1}{2^(n-1)}=3)
(故S_{n}是递增有界数列,可知其必有极限,设其极限为S)
(则S=lim_{n o infty}frac{1}{0!}+frac{1}{1!}+frac{1}{2!}+cdotcdotcdot+frac{1}{n!})
(对e_{n}进行二项式展开)
(e_{n}={(1+frac{1}{n})}^{n}) (其中,n(in)(N^{+}))
(quad=sum_{k=0}^{n})(C_{n}^{n-k})((frac{1}{n})^{k})
(quad=1+sum_{k=1}^{n})(C_{n}^{n-k})(frac{1}{n^{k}})
(quad=1+sum_{k=1}^{n}frac{n!}{(n-k)!k!}frac{1}{n^{k}})
(quad=1+sum_{k=1}^{n}frac{1}{k!}frac{n!}{(n-k)!}frac{1}{n^{k}})
(quad=1+sum_{k=1}^{n}frac{1}{k!}frac{(n-k+1)(n-k+2)cdotcdotcdot n}{n^{k}})
(因为 1*2*3cdotcdotcdot(n-k)(n-k+1)(n-k+2)...n)
从1到n-k,一共是n-k个连续数字相乘,从n-k+1到n,合计k个连续数字相乘,从1到n,合计是n个连续数字相乘
故
上式(=1+sum_{k=1}^{n}frac{1}{k!}frac{(n-k+1)(n-k+2)(n-k+3)cdotcdotcdot (n-2)(n-1)n(一共k个数字)}{nnncdotcdotcdot n(一共k个n)})
(quad=1+sum_{k=1}^{n}frac{1}{k!}frac{n(n-1)(n-2)cdotcdotcdot (n-k+3)(n-k+2)(n-k+1)(一共k个数字)}{nnncdotcdotcdot n(一共k个n)})
(quad=1+sum_{k=1}^{n}frac{1}{k!}frac{n}{n}frac{n-1}{n}frac{n-2}{n}cdotcdotcdotfrac{n-k+2}{n}frac{n-k+1}{n})
(quad=1+sum_{k=1}^{n}frac{1}{k!}*1*(1-frac{1}{n})(1-frac{2}{n})cdotcdotcdotfrac{n-k+2}{n}frac{n-k+1}{n})
(quad=1+sum_{k=1}^{n}frac{1}{k!}(1-frac{1}{n})(1-frac{2}{n})cdotcdotcdot (1-frac{k-2}{n})(1-frac{k-1}{n})) (一共k-1个括号)
展开连加号
(quad=1+frac{1}{1!}+frac{1}{2!}(1-frac{1}{n})+frac{1}{3!}(1-frac{1}{n})(1-frac{2}{n})+cdotcdotcdot+frac{1}{n!}(1-frac{1}{n})(1-frac{2}{n})cdotcdotcdot(1-frac{n-1}{n}))
上式最后一项,是取k=n,一共n-1个括号
上式一共n+1项
(由上式可知\)
(e_{n}leqslant1+frac{1}{1!}+frac{1}{2!}+frac{1}{3!}+cdotcdotcdot+frac{1}{n!})
(quadleqslant1+1+frac{1}{2}+frac{1}{2^2}+frac{1}{2^3}cdotcdotcdot+frac{1}{2^n})
(=1+1*frac{1-frac{1}{2}^n}{1-frac{1}{2}})
(=1+2*(1-frac{1}{2}^n))
(=1+2-frac{1}{2^{n-1}})(\)
< 3
(且e_{n}leqslant S)
(e_{n+1}={(1+frac{1}{n+1})}^{n+1}) (其中,n(in)(N^{+}))
(quad=sum_{k=0}^{n+1})(C_{n+1}^{n+1-k})((frac{1}{n+1})^{k})
(quad=sum_{k=0}^{n+1}frac{1}{k!}frac{(n+1)!}{(n+1-k)!}frac{1}{(n+1)^k})
(quad=sum_{k=0}^{n+1}frac{1}{k!}frac{(n+1-k+1)(n+1-k+2)(n+1-k+3)cdotcdotcdot(n+1)(k个括号)}{(n+1)^k}) bbbb
(quad=sum_{k=0}^{n+1}frac{1}{k!}frac{(n+1-k+1)(n+1-k+2)(n+1-k+3)cdotcdotcdot(n+1)(k个括号)}{(n+1)cdotcdotcdot(n+1)(k个(n+1))})
(quad=sum_{k=0}^{n+1}frac{1}{k!}frac{(n+1)n(n-1)cdotcdotcdot(n+1-k+3)(n+1-k+2)(n+1-k+1)(k个括号)}{(n+1)cdotcdotcdot(n+1)(k个(n+1))})
(quad=1+sum_{k=1}^{n+1}frac{1}{k!}(1-frac{1}{n+1})(1-frac{2}{n+1})cdotcdotcdot (1-frac{k-2}{n+1})(1-frac{k-1}
{n+1})) (一共k-1个括号)(\)
(quad=1+frac{1}{1!}+frac{1}{2!}(1-frac{1}{n+1})+frac{1}{3!}(1-frac{1}{n+1})(1-frac{2}{n+1})+cdotcdotcdot+frac{1}{(n+1)!}(1-frac{1}{n+1})(1-frac{2}{n+1})cdotcdotcdot(1-frac{n}{n+1}))
(即:e_{n}=1+frac{1}{1!}+frac{1}{2!}(1-frac{1}{n})+frac{1}{3!}(1-frac{1}{n})(1-frac{2}{n})+cdotcdotcdot+frac{1}{n!}(1-frac{1}{n})(1-frac{2}{n})cdotcdotcdot(1-frac{n-1}{n})\)
(即:e_{n+1}=1+frac{1}{1!}+frac{1}{2!}(1-frac{1}{n+1})+frac{1}{3!}(1-frac{1}{n+1})(1-frac{2}{n+1})+cdotcdotcdot+frac{1}{(n+1)!}(1-frac{1}{n+1})(1-frac{2}{n+1})cdotcdotcdot(1-frac{n}{n+1})\)
(可知e_{n+1}为n+2项,e_{n}为n+1项,e_{n+1}比e_{n}多一项,且前面的n+1项都大于e_{n}的对应位置的项\)
(可知e_{n+1}>e_{n}, 可知e_{n}为递增数列,且有上界3,根据单调递增有界数列必有极限,可知e_{n}有极限。)(\)
(为e,即lim_{n o infty}e_{n}=e)
(即lim_{n o infty}e_{n}=lim_{n o infty}(1+frac{1}{1!}+frac{1}{2!}(1-frac{1}{n})+frac{1}{3!}(1-frac{1}{n})(1-frac{2}{n})+cdotcdotcdot+frac{1}{n!}(1-frac{1}{n})(1-frac{2}{n})cdotcdotcdot(1-frac{n-1}{n})))
(forall min N^+且mleqslant n,设\)
(e_{n,m}=1+frac{1}{1!}+frac{1}{2!}(1-frac{1}{n})+cdotcdot+frac{1}{m!}(1-frac{1}{n})(1-frac{2}{n})cdotcdotcdot(1-frac{m-1}{n})\)
(即:e_{n,m}是e_{n}的前m项和,所以,forall n 都有下面的不等式成立)
(e_{n}geqslant e_{n,m})
(根据数列极限的保不等式性,两侧对n取极限,不等式依然成立,即:\)
(lim_{n o infty}e_{n}geqslant lim_{n o infty}e_{n,m}=1+1+frac{1}{2!}+frac{1}{3!}+cdotcdot+frac{1}{m!})
(即quad e geqslant lim_{n o +infty}e_{n,m}=1+1+frac{1}{2!}+frac{1}{3!}+cdotcdot+frac{1}{m!})
(此时该不等式左侧为常量e,右侧的最终结果,已经不包含变量n,仅包含变量m,而m的要求是mleqslant n,此时n为无穷大,\)
(所以m可以取任意值,即forall m,都有 egeqslant lim_{n o infty}e_{n,m}=1+1+frac{1}{2!}+frac{1}{3!}+cdotcdot+frac{1}{m!}=S_{m}\)
由数列极限的保不等式性,对m取极限,可得
(egeqslant lim_{m o infty}lim_{n o infty}e_{n,m}=lim_{m o infty}frac{1}{0!}+frac{1}{1!}+frac{1}{2!}+cdotcdot++frac{1}{m!}=S)
(而前面已经证明 eleqslant S)
(故,得到Sleqslant eleqslant Squadquad (注意 geqslant意为"不小于",leqslant意为“不大于”))
所以 e=S,即
(e=lim_{n o infty}(frac{1}{0!}+frac{1}{1!}+frac{1}{2!}+cdotcdot+...frac{1}{n!}))
(说明:在n o infty的过程中,e_{n}各项都在增大,趋于对应的阶乘倒数,\)
(在n取无穷大时,e_{n o +infty}所有项的极限都是阶乘倒数,其极限和的极限是倒数阶乘之和\)