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一道有难度的极限题

陈洪葛 posted @ 13 年前 in 数学分析 , 1418 阅读

计算:

lim

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}.


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