命题1. 设
\[a_n=\left(1+\frac1n\right)^n,\quad b_n=\frac{1}{0!}+\frac{1}{1!}+\cdots+\frac{1}{n!}=\sum_{k=0}^\infty \frac{1}{k!},\]
则\(\{a_n\}\), \(\{b_n\}\)均收敛并且\(\lim\limits_{n\to \infty}a_n=\lim\limits_{n\to \infty}b_n\).
证明：
Step1. 显然\(\{b_n\}\)是严格增数列. 计算可知\(a_1=b_1=2\), \(a_1<a_2\). 对任意\(n\geq 2\), 有
\[
\begin{array}{rcl}
a_n&=&\left(1+\frac1n\right)^n\\
&=&1+1+C_n^2\cdot \frac{1}{n^2}+\cdots +C_n^k \cdot \frac{1}{n^k}+\cdots +\frac{1}{n^n}\\
&=&2+\sum\limits_{k=2}^{n}C_n^k \cdot\frac{1}{n^k}\\
&=&2+\sum\limits_{k=2}^{n}\frac{n(n-1)(n-2)\cdot (n-k+1)}{k!}\cdot \frac{1}{n^k}\\
&=&2+\sum\limits_{k=2}^n \frac{1}{k!}\cdot \frac{n}{n}\cdot \frac{n-1}{n}\cdot \cdots \cdot \frac{n-k+1}{n}\\
&=&2+ \sum\limits_{k=2}^n \frac{1}{k!}\left(1-\frac1n\right)\left(1-\frac{2}{n}\right)\cdots\left( 1-\frac{k-1}{n}\right)\tag{1}\\
&<&2+\sum\limits_{k=2}^{n+1} \frac{1}{k!}\left(1-\frac{1}{n+1}\right)\left(1-\frac{2}{n+1}\right)\cdots\left( 1-\frac{k-1}{n+1}\right)\\
&=&2+\sum\limits_{k=2}^{n+1} C_{n+1}^k \cdot \frac{1}{n^k}\\
&=&a_{n+1},
\end{array}
\]
所以\(\{a_n\}\)是严格增数列. 另一方面, 由(1)式可得
\[a_n\leq 2+\sum_{k=2}^{n}\frac{1}{k!}=b_n,\quad \forall n\geq 2.\tag{2}\]
对任意\(n\geq 2\), 有
\[\begin{array}{rcl}
b_n&=&2+\frac{1}{2!}+\frac{2}{3!}+\cdots+\frac{1}{n!}\\
&<&2+\frac{1}{1\cdot 2}+\frac{!}{2\cdot 3}+\cdots +\frac{1}{(n-1)\cdot n}\\
&=&2+1-\frac1n\\
&<&3.\tag{3}
\end{array}\]
综合(1)-(3)式, \(\{a_n\}\)和\(\{b_n\}\)都是严格增数列并且
\[a_n\leq b_n<3,\quad \forall n\in \Bbb{N}_+. \tag{4}\]
根据单调有界定理, \(\{a_n\}\)和\(\{b_n\}\)都收敛.
Step2. 设\(A=\lim\limits_{n\to \infty}a_n\), \(B=\lim\limits_{n\to \infty}b_n\). 根据(4)式, 利用数列极限的保不等式性可得\(A\leq B\). 下证\(A=B\).
任取\(n\in \Bbb{N}_+\)且\(n\geq 2\), 固定\(n\). 对任意\(m>n\), 都有
\[\begin{array}{rcl}
a_m&=&2+\sum\limits_{k=2}^m \frac{1}{k!}\left(1-\frac1m\right)\left(1-\frac{2}{m}\right)\cdots\left( 1-\frac{k-1}{m}\right)\\
&>&2+\sum\limits_{k=2}^n \frac{1}{k!}\left(1-\frac1m\right)\left(1-\frac{2}{m}\right)\cdots\left( 1-\frac{k-1}{m}\right)
\end{array}\]
在上式两端令\(m\to \infty\), 根据数列极限的保不等式性可得
\[\begin{array}{rcl}
&&A=\lim\limits_{m\to \infty}a_m\\
&\geq& \lim\limits_{m\to \infty}\left[2+\sum\limits_{k=2}^n \frac{1}{k!}\left(1-\frac1m\right)\left(1-\frac{2}{m}\right)\cdots\left( 1-\frac{k-1}{m}\right)\right]\\
&=&2+\sum\limits_{k=2}^n \lim\limits_{m\to \infty}\left[ \frac{1}{k!}\left(1-\frac1m\right)\left(1-\frac{2}{m}\right)\cdots\left( 1-\frac{k-1}{m}\right)\right]\\
&=&2+\sum\limits_{k=2}^n \frac{1}{k!}=b_n,\quad \forall n\geq 2,
\end{array}\]
从而
\[A\geq \lim_{n\to \infty}b_n=B.\]
综上\(A=B\).
\(\Box\)
根据\(\{a_n\}\)和\(\{b_n\}\)的严格增性以及(4)式, 易证
\[0<2=a_1\leq a_n\leq b_n<A=B<3,\quad \forall n\in \Bbb{N}_+.、\tag{5}\]
定义1. 记
\[e=\lim_{n\to \infty}\left(1+\frac1n\right)^n=\lim_{n\to \infty}\left(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\cdots+\frac{1}{n!}\right),\]
称以\(e\)为底数的对数为自然对数, 记为
\[\ln x=\log_e x,\quad x>0.\]
推论1. (误差估计) 令\(b_n=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\cdots+\frac{1}{n!}=\sum\limits_{k=0}^n \frac{1}{k!}\), 则
\[ 0<e-b_n<\frac{1}{n!n},\quad \forall n\in \Bbb{N}_+.\]
证明:
对任意\(m,n\in \Bbb{N}_+\), 有
\[\begin{array}{rcl}
&&b_{m+n}-b_n\\
&=&\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\cdots +\frac{1}{(m+n)!}\\
&=&\frac{1}{(n+1)!}\left[1+\frac{1}{n+2}+\frac{1}{(n+2)(n+3)}+\cdots+\frac{1}{(n+2)(n+3)\cdots(n+m)} \right]\\
&<&\frac{1}{(n+1)!}\left[1+\frac{1}{n+1}+\frac{1}{(n+1)^2}+\cdots +\frac{1}{(n+1)^{m-1}} \right]\\
&=&\frac{1}{(n+1)!}\cdot \frac{1-\frac{1}{(n+1)^m}}{1-\frac{1}{n+1}}\\
&<&\frac{1}{(n+1)!}\cdot \frac{1}{1-\frac{1}{n+1}}\\
&=&\frac{1}{n!n},
\end{array}\]
在上式两端令\(m\to \infty\)可得
\[0<e-b_n\leq \frac{1}{n!n}.\]
\(\Box\)
推论2. \(e\)是无理数.
证明:
由(5)式可知, \(2<e<3\), 所以\(e\)不是整数.
反证法, 假设\(e\)是介于2和3之间的有理数,
\[e=\frac{p}{q},\]
其中\(p,q\in \Bbb{N}_+\)且\(p,q\)互质. 由于\(e\in (2,3)\), 则
\[p\geq 2,\quad q\geq 2.\]
由推论1可知,
\[0<e-b_q\leq \frac{1}{q!q},\]
于是
\[0<q!(e-b_q)\leq \frac1q\leq \frac12,\]
\(q!(e-b_q)\)不是整数. 但是另一方面,
\[\begin{array}{rcl}
q!(e-b_q)&=&q!\left[\frac{p}{q}-\left(2+\frac{1}{2!}+\cdots+\frac{1}{q!}\right)\right]\\
&=&(q-1)!p-q!\left(2+\frac{1}{2!}+\cdots+\frac{1}{q!}\right),
\end{array}\]
上式最右端为整数, 从而\(q!(e-b_q)\)是整数, 矛盾.
\(\Box\)
命题2. 设
\[c_n=\left(1+\frac1n \right)^{n+1},\]
则\(\{c_n\}\)严格减并且
\[\left(1+\frac1n\right)^n<e<\left(1+\frac1n\right)^{n+1},\tag{6}\]
\[\lim_{n\to \infty} \left(1+\frac1n\right)^{n+1}=e.\]
证明:
对任意\(n\in \Bbb{N}_+\),
\[\begin{array}{rcl}
\frac{c_n}{c_{n+1}}&=& \frac{\left(1+\frac1n\right)^{n+1}}{ \left(1+\frac{1}{n+1}\right)^{n+2}}\\
&=&\left(\frac{1+\frac{1}{n}}{1+\frac{1}{n+1}}\right)^{n+1}\cdot \frac{1}{1+\frac{1}{n+1}}\\
&=&\left(1+\frac{1}{n^2+2n}\right)^{n+1}\cdot \frac{n+1}{n+2}\\
&>&\left(1+\frac{n+1}{n^2+2n}\right)\cdot \frac{n+1}{n+2}\\
&>&\left(1+\frac{1}{n+1}\right) \cdot \frac{n+1}{n+2}\\
&=&1,
\end{array}\]
所以\(\{c_n\}\)严格减.
由于\(c_n=a_n\left(1+\frac1n\right)\), 则
\[\lim_{n\to \infty} c_n =\lim_{n\to \infty}a_n \cdot \lim_{n\to \infty}\left(1+\frac1n\right)=e,\]
从而\(e=\inf\limits_{n}c_n\). 根据\(\{c_n\}\)的严格减性, 就有
\[e<c_n,\quad \forall n\in \Bbb{N}_+.\]
\(\Box\)
对(6)式取自然对数可以得到以下结论.
推论3. \[\frac{1}{n+1}<\ln \left(1+\frac{1}{x}\right)<\frac1n,\quad \forall n\in \Bbb{N}_+.\]
推论4. 设
\[d_n=1+\frac12 +\frac13+\cdots+\frac1n -\ln n,\]
则\(\{d_n\}\)收敛.
证明:
由推论3,
\[\begin{array}{rcl}
d_{n+1}-d_n&=&\frac{1}{n+1}-\ln (n+1) +\ln n\\
&=&\frac{1}{n+1}-\ln\left(1+\frac1n\right)<0,
\end{array}\]
所以\(\{d_n\}\)严格减. 另一方面,
\[\begin{array}{rcl}
d_n&=&1+\frac12 +\cdots +\frac1n -\ln n\\
&=&1+\frac12 +\cdots +\frac1n-\ln \left(\frac{n}{n-1}\cdot \frac{n-1}{n-2}\cdots \frac32\cdot \frac21 \right)\\
&=&\sum\limits_{k=1}^n \frac1k -\sum\limits_{k=1}^{n-1}\ln \left(1+\frac{1}{k}\right)\\
&=&\sum\limits_{k=1}^{n-1} \left[\frac{1}{k}-\ln\left(1+\frac1k\right) \right]+\frac1n\\
&>&\frac1n>0,
\end{array}\]
所以\(\{d_n\}\)有下界. 根据单调有界定理, \(\{d_n\}\)收敛.
\(\Box\)
定义2. 称
\[\gamma=\lim_{n\to \infty}d_n=\lim_{n\to \infty}\left(1+\frac12 +\frac13+\cdots+\frac1n -\ln n\right)\]
为Euler常数.
注: 由于\(d_n>0\) (\(\forall n\in\Bbb{N}_+\)), 再根据\(\{d_n\}\)的严格减性可知\(\gamma>0\). 通过计算可得
\[\gamma=0.5771\cdots.\]
例. 证明\(S_n=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n}\)收敛, 并求出极限.
证明:
设
\[d_n=1+\frac12 +\frac13+\cdots+\frac1n -\ln n,\]
则
\[d_{2n}-d_n=S_n-\ln{(2n)}+\ln n=S_n-\ln 2.\]
由于
\[\lim_{n\to \infty}d_{2n}=\lim_{n\to \infty}d_n=\gamma,\]
则
\[\lim_{n\to \infty}S_n=\ln 2.\]
问题. \(e\)和圆周率\(\pi\)都是无理数, 那么\(e-\pi\)和\(e+\pi\)是否是无理数？到目前为止, 这一问题仍未解决. Euler常数\(\gamma\)是否是无理数这一问题也没有证明.