计算：

\[ \lim_{n\rightarrow\infty}{\left(1-\frac{1}{1\cdot 2}\right)\left(1-\frac{1}{2\cdot 3}\right)\cdots\left(1-\frac{1}{n(n+1)}\right)} \]

Solution：

We write

$ P=\prod_{n=1}^\infty \left(1-\frac{1}{n(n+1)}\right)=\prod_{n=1}^\infty \left(\frac{n^2+n-1}{n^2+n}\right)=\prod_{n=1}^\infty\frac{(n-a_1)(n-a_2)}{(n-b_1)(n-b_2)}  $,

where  $a_1=\frac{-1+\sqrt{5}}{2},\;a_2=\frac{-1-\sqrt{5}}{2},\;b_1=0$, and $b_2=-1.$

Using the [i]Weierstrass product[/i] for the gamma function, $\frac{1}{\Gamma(x)}=xe^\gamma x\prod_{n=1}^\infty \left(1+\frac{x}{n} \right)e^{-x/n}$, where $\gamma$ denotes Euler's constant, and the fundamental relation $\Gamma(x+1)=x\Gamma(x),$ we can deduce that

 $P= \prod_{j=1}^2 \frac{\Gamma(1-b_j)}{\Gamma(1-a_j)}=\frac{\Gamma(1)\Gamma(2)}{\Gamma\left(\frac{3-\sqrt{5}}{2}\right)\Gamma\left(\frac{3+\sqrt{5}}{2}\right)}=\frac{1}{\Gamma\left(\frac{3-\sqrt{5}}{2}\right)\Gamma\left(\frac{3+\sqrt{5}}{2}\right)}\approx 0.2966751356.$

[This standard method of determining certain infinite products is explained, for example, in Section 12.13 of [i]A Course of Modern Analysis [/i]([i]Fourth Edition[/i]) by E. T. Whittaker and G. N. Watson.]

Using the relation $\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin\pi x},$ for $x$ not an integer, it should be possible to deduce a closed form for the infinite product given by the computer algebra system [i]Maple[/i]:  $\; - \frac{\sin \pi\left(\frac{1+\sqrt{5}}{2}\right)}{\pi}.$