# Evaluating $\prod\limits_{n=2}^{\infty}\left(1-\frac{1}{n^3}\right)$

I'm having a very hard time understanding the following evaluation of $\displaystyle\prod_{n=2}^{\infty}\left(1-\frac{1}{n^3}\right)$.

The beginning and end make sense, but I can't make much sense out of the middle.

Here $\omega = -\frac{1}{2}+ i \frac{\sqrt{3}}{2}$ (a primitive third root of unity).

$$\prod_{n=2}^{\infty}\left(1-\frac{1}{n^3}\right) =\prod_{n=2}^{\infty}\frac{(n-1)(n^2+n+1)}{n^3}$$

$$=\lim_{m\to\infty}\frac{1}{m}\prod_{n=2}^m\frac{(n-\omega)(n-\omega^2)}{n^2}$$

$$=\lim_{m\to\infty}\frac{\Gamma(m+1-\omega)\Gamma(m+1-\omega^2)}{m(m!)^2\Gamma(-\omega)\Gamma(-\omega^2)(1-\omega)(1-\omega^2)(-\omega)(-\omega^2)}$$

$$=\frac{1}{3\Gamma(-\omega)\Gamma(-\omega^2)}=\frac{\sin{\pi(-\omega)}}{3\pi}$$

$$=\frac{\cosh (\frac{\sqrt{3}\pi}{2})}{3\pi }$$

EDIT: I think I'm starting to make sense out of this by writing out the terms.

$$\prod_{n=2}^{m} \frac{(n- \omega)(n-\omega^{2})}{n^{2}}= \frac{1}{2^{2} \times 3^{2} \cdots m^{2}} (2-\omega)(3-\omega) \cdots (m- \omega)(2-\omega^{2})(3-\omega^{2})$$

$$\cdots (m- \omega^{2})$$

$$= \frac{1}{(m!)^{2}} \frac{\Gamma(m+1-\omega)}{(1-\omega)(-\omega)\Gamma(-\omega)} \frac{\Gamma(m+1-\omega^{2})}{(1-\omega^{2})(-\omega^{2})\Gamma(-\omega^{2})}$$

Now I have to figure out how to take the limit.

• What part of the middle? – Qiaochu Yuan Mar 3 '13 at 20:36
• line two to line three – Random Variable Mar 3 '13 at 20:38
• The $\frac{1}{m}$ comes from the evaluation of $\prod_{n=2}^{m} \frac{n-1}{n}$. – Random Variable Mar 3 '13 at 20:49
• @RandomVariable Please consider posting the solution (or its sketch) as an answer. – Sasha Mar 3 '13 at 21:26
• @RandomVariable: math.stackexchange.com/questions/168740/… – user 1357113 Mar 3 '13 at 21:59

OK. Here's my understanding of that evaluation.

$$\prod_{n=2}^{\infty}\left(1-\frac{1}{n^3}\right) = \prod_{n=2}^{\infty}\frac{(n-1)(n^2+n+1)}{n^3}$$

$$= \prod_{n=2}^{\infty} \frac{(n-1)(n- \omega)(n-\omega^{2})}{n^{3}}$$

$$= \lim_{m \to \infty} \prod_{n=2}^{m} \frac{n-1}{n} \frac{(n- \omega)(n-\omega^{2})}{n^{2}}$$

$$= \lim_{m \to \infty} \frac{1}{2} \cdot \frac{2}{3}\ \cdots \frac{m-1}{m} \frac{1}{2^{2} \cdot 3^{2} \cdots m^{2}} (2-\omega)(3-\omega) \cdots (m- \omega) (2-\omega^{2})(3-\omega^{2})$$$$\cdots (m- \omega^{2})$$

$$= \lim_{m \to \infty} \frac{1}{m} \frac{1}{(m!)^{2}} \displaystyle \frac{\Gamma(m+1-\omega)}{(1-\omega)(-\omega)\Gamma(-\omega)} \frac{\Gamma(m+1-\omega^{2})}{(1-\omega^{2})(-\omega^{2})\Gamma(-\omega^{2})}$$

And since $(1-\omega)(1-\omega^{2}) = 3$ and $(-\omega)(-\omega^{2})=1$,

$$= \frac{1}{3\Gamma(-\omega) \Gamma(-\omega^{2})}\lim_{m \to \infty} \frac{\Gamma(m+1- \omega) \Gamma(m+1-\omega^{2})}{m \Gamma(m+1) \Gamma(m+1) }$$

That limit evaluates to one according to Wolfram Alpha, but I don't know why. (See below)

$$= \frac{1}{3 \Gamma \left(\frac{1}{2} - i\frac{\sqrt{3}}{2} \right) \Gamma \left(\frac{1}{2} + i\frac{\sqrt{3}}{2} \right)} = \frac{1}{3 \Gamma \left(1- \frac{1+ i \sqrt{3}}{2} \right) \Gamma \left(\frac{1}{2} + i\frac{\sqrt{3}}{2} \right)}$$

$$= \frac{\sin \left(\pi \frac{1+i \sqrt{3}}{2} \right)}{3 \pi}= \frac{\sin \frac{\pi}{2} \cos \frac{i \pi \sqrt{3}}{2} + \cos \frac{\pi}{2} \sin \frac{i \pi \sqrt{3}}{2}}{3 \pi}$$

$$= \frac{\cosh \frac{\pi \sqrt{3}}{2}}{3 \pi}$$

EDIT:

According to Wikipedia, $\displaystyle \lim_{n \to \infty} \frac{\Gamma (n+a)}{\Gamma(n)n^{a}} = 1$. Wikipedia says it's valid for $a \in \mathbb{R}$. But I'm pretty sure it's valid for complex $a$ as well.

http://en.wikipedia.org/wiki/Gamma_function#General

Assuming that it is,

$$\lim_{m \to \infty} \frac{\Gamma(m+1- \omega) \Gamma(m+1-\omega^{2})}{m \Gamma(m+1) \Gamma(m+1) }$$

$$= \lim_{m \to \infty} \frac{\Gamma(m+1- \omega) \Gamma(m+1-\omega^{2})}{m^{-\omega}m^{-\omega^{2}} \Gamma(m+1) \Gamma(m+1) }$$

$$= \lim_{m \to \infty} \frac{(m-\omega)(m-\omega^{2})}{m^{2}}\frac{\Gamma(m- \omega)}{\Gamma(m) m^{-\omega} } \frac{\Gamma(m-\omega^{2})}{ \Gamma(m)m^{-\omega^{2}} } = 1$$

SECOND EDIT:

For a general way to evaluate $\displaystyle \prod_{n=N}^{\infty} \left(1- \frac{1}{n^{k}} \right)$, see sos440's post at the following link:

http://integralsandseries.prophpbb.com/topic221.html#p1560

• Stirling formula might help. – Sungjin Kim May 30 '13 at 20:15