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I have a trouble with this limit of the infinite product: $$\lim _{n \to\infty}\left(1-\frac{1}{1 \cdot 2}\right)\left(1-\frac{1}{2 \cdot 3}\right) \cdots\left(1-\frac{1}{n(n+1)}\right)$$ My attempt:

We have $$\prod_{n=1}^{\infty}\left(1-\frac{1}{n(n+1)}\right)=\prod_{n=1}^{\infty}\left(\frac{n^{2}+n-1}{n^{2}+n}\right)=\prod_{n=1}^{\infty} \frac{\left(n-a_{1}\right)\left(n-a_{2}\right)}{n \left(n+1\right)},$$ where $a_{1}=\dfrac{-1+\sqrt{5}}{2}$, $a_{2}=\dfrac{-1-\sqrt{5}}{2}$.

So I would just like a hint as to how to proceed. Any help would be appreciated.

Anne Bauval
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unicornki
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2 Answers2

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I continue from where you left. The infinite product representation of the sine function can be used to finish the calculation: \begin{align*} & \mathop {\lim }\limits_{N \to + \infty } \prod\limits_{n = 1}^N {\left( {1 - \frac{{a_1 }}{n}} \right)\left( {1 + \frac{{ - 1 - a_2 }}{{n + 1}}} \right)} = \mathop {\lim }\limits_{N \to + \infty } \prod\limits_{n = 1}^N {\left( {1 - \frac{{a_1 }}{n}} \right)\left( {1 + \frac{{a_1 }}{{n + 1}}} \right)} \\ & = \mathop {\lim }\limits_{N \to + \infty } \frac{{1 + \frac{{a_1 }}{{N + 1}}}}{{1 + a_1 }}\prod\limits_{n = 1}^N {\left( {1 - \frac{{a_1 }}{n}} \right)\left( {1 + \frac{{a_1 }}{n}} \right)} = \frac{1}{{1 + a_1 }}\prod\limits_{n = 1}^\infty {\bigg( 1 - \frac{{a_1^2 }}{{n^2 }} \bigg)} \\ & = \frac{1}{{1 + a_1 }}\frac{{\sin (\pi a_1 )}}{{\pi a_1 }} = \frac{{\sin (\pi a_1 )}}{{\pi }}. \end{align*}

Gary
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By means of Weierstrass product of the gamma function，observe that $$\frac{1}{\Gamma(z)}=e^{\gamma z} z \prod_{n=1}^{\infty}\left(1+\frac{z}{n}\right) e^{-z / n}$$ where $\gamma $ is the Euler–Mascheroni constant. Then $$\prod_{n=1}^{\infty} \frac{\left(n-a_{1}\right)\left(n-a_{2}\right)}{n(n+1)}=\frac{\Gamma\left(1\right)\Gamma\left(2\right)}{\Gamma\left(1-a_{1}\right)\Gamma\left(1-a_{2}\right)}=\frac{\Gamma(1) \Gamma(2)}{\Gamma\left(\frac{3-\sqrt{5}}{2}\right) \Gamma\left(\frac{3+\sqrt{5}}{2}\right)}.$$ Since $\Gamma(x+1)=x \Gamma(x)$ and $\Gamma(n)=(n-1) !$ for any positive integer $n$, $$\frac{\Gamma(1) \Gamma(2)}{\Gamma\left(\frac{3-\sqrt{5}}{ 2}\right) \Gamma\left(\frac{3+\sqrt{5}}{2}\right)}=\frac{1}{\Gamma\left(\frac{3-\sqrt{5}}{2}\right) \Gamma\left(\frac{3+\sqrt{5}}{2}\right)}.$$ Using the relation $\Gamma(x) \Gamma(1-x)=\frac{\pi}{\sin \pi x}$, thus we obtain $$\frac{1}{\Gamma\left(\frac{3-\sqrt{5}}{2}\right) \Gamma\left(\frac{3+\sqrt{5}}{2}\right)}=\frac{1}{(\frac{1-\sqrt{5}}{2})(\frac{1+\sqrt{5}}{2})}\frac{1}{\Gamma\left(\frac{1-\sqrt{5}}{2}\right) \Gamma\left(\frac{1+\sqrt{5}}{2}\right)}=-\frac{\sin \left(\frac{(1+\sqrt{5})\pi}{2}\right)}{\pi}$$

unicornki
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