先给出结论：

对于数列 \({\varphi_{n}},n\in\mathbf{N}^{+}\)，若有： $$ \sum_{i=1}^\infty\varphi_{i}=\beta>1,\varphi_{i}\in(0,1)$$ 那么： $$(1-\varphi_{2})(1-\varphi_{3})(1-\varphi_{4})\cdots(1-\varphi_{n})>1+\varphi_{1}-\beta\qquad(1)$$ $$(1+\varphi_{2})(1+\varphi_{3})(1+\varphi_{4})\cdots(1+\varphi_{n})<\frac{1}{1+\varphi_{1}-\beta}\qquad(2)$$

下面我们先用数学归纳法来证明(1)式：

1. 当\(n=2\)时，\(1-\varphi_{2}-(1+\varphi_{1}-\beta) = \beta - \varphi_{1} - \varphi_{2}>0\)，命题成立。
2. 当\(n\geq3\)时，不妨设$$p_{n}=\prod_{i=2}^{n}(1-\varphi_{i})$$则\(p_{2}=1-\varphi_{2}\)，易得\(p_{n}<1\)。 $$\therefore p_{n} = (1-\varphi_{n})p_{n-1} = p_{n-1}-p_{n-1}\varphi_{n} = p_{n-1}-\varphi_{n}\;(n \geq 2)$$ $$\therefore p_{n}>p_{n-1}-\varphi_{n}>p_{n-2}-(\varphi_{n-1}+\varphi_{n})>\cdots>p_{2}-\sum_{i=3}^{n}\varphi_{i}= 1-\sum_{i=2}^{n}\varphi_{i}>1-\sum_{i=2}^{\infty}\varphi_{i}=1+\varphi_{1}-\beta$$ $$\therefore p_{n}=(1-\varphi_{2})(1-\varphi_{3})\cdots(1-\varphi_{n})>1+\varphi_{1}-\beta$$
3. 至此，(1)式证毕。

下面我们来证明(2)式：

再设： $$p_{n}=\prod_{i=2}^{n}(1-\varphi_{i})，t_{n}=\prod_{i=2}^{n}(1+\varphi_{i})$$ 则有： $$k_{n}=p_{n}\cdot t_{n}=\prod_{i=2}^{n}(1-\varphi_{i}^{2})\quad(n\in\mathbf{N}^{+})$$ 又由(1)中结论得知\(p_{n}>(1-\varphi_{1})(1+\varphi_{1}-\beta)>0\)，而\(k_{n}<1\)， $$\therefore t_{n}= \frac{k_{n}}{p_{n}}<\frac{1}{(1-\varphi_{1})(1+\varphi_{1}-\beta)}$$故\(t_{n}\)必有上限。

设$$\lim_{n \to \infty}t_{n}=C$$ 则易知\(t_{n}<C\)，\(t_{1}=1+\varphi_{1}\)，\(t_{n}=t_{n-1}\cdot(1+\varphi_{n})\quad(n\geq 2)\)

下面用数学归纳法继续证明(2)成立：

1. 当\(n=2\)时， $$\because (1+\varphi_{2})-\frac{1}{1+\varphi_{1}-\beta}>1+\varphi_{2}-\frac{1}{1-\varphi_{2}}=\frac{-\varphi_{2}^{2}}{1-\varphi_{2}}<0$$ $$\therefore 1+\varphi_{2}<\frac{1}{1+\varphi_{1}-\beta}$$
2. 当\(n\geq3\)时，\(t_{n}<t_{n-1}+\varphi_{n}\cdot C\)， $$\therefore t_{n}<t_{n-1}+\varphi_{n}\cdot C<t_{n-2}+(\varphi_{n-1}+\varphi_{n})\cdot C<\cdots <t_{1}+C\cdot \sum_{i=2}^{\infty}\varphi_{i}<1+\varphi_{1}+C\cdot(\beta-\varphi_{1}),$$ \(\therefore\)当\(n\to\infty\)时，\(C\leq1+\varphi_{1}+C\cdot(\beta-\varphi_{1})\)，可解得\(C\leq\frac{1+\varphi_{1}}{1+\varphi_{1}-\beta}\)， \(\therefore t_{n}<\frac{1+\varphi_{1}}{1+\varphi_{1}-\beta}\)， $$\therefore (1+\varphi_{1})(1+\varphi_{2})\cdots(1+\varphi_{n})<\frac{1}{1+\varphi_{1}-\beta}$$
3. 这样我们就证明了(2)。

由此我们可以获得一些有趣的结论（括号中是该结论所用到的极限）：

$$\prod_{i=2}^{\infty}\big(1-\frac{1}{i!}\big)>\frac{7}{4}-\frac{e}{2},\quad\prod_{i=2}^{\infty}\big(1+\frac{1}{i!}\big)<\frac{3}{7-2e}\qquad\bigg(\sum_{i=2}^{\infty}{\frac{1}{i!}}=e-2\bigg)$$

$$\prod_{i=2}^{\infty}\big(1-\frac{1}{i^{2}}\big)>\frac{27}{16}-\frac{\pi^{2}}{8},\quad\prod_{i=2}^{\infty}\big(1+\frac{1}{i^{2}}\big)<\frac{15}{27-2\pi^{2}}\qquad\bigg(\sum_{i=2}^{\infty}{\frac{1}{i^{2}}}=\frac{\pi^{2}}{6}-1\bigg)$$

$$\prod_{i=2}^{\infty}(1-C_{n}^{i}\cdot\frac{1}{i!})>\frac{7}{4}-\frac{e}{2},\quad\prod_{i=2}^{\infty}(1+C_{n}^{i}\cdot\frac{1}{i!})<\frac{3}{7-2e}\Bigg(\lim_{n \to \infty}\bigg(\big(\frac{1}{n}+1\big)^{n}-2\bigg)=\lim_{n \to \infty}\bigg(\sum_{i=2}^{n}C_{n}^{i}\cdot\frac{1}{n^{i}}\bigg)=e-2\Bigg)$$

2012-10-21 全文完