Evaluating the infinite product $\prod_{k\geq 2}(1-\frac{1}{k^3})$

Question

Does anyone know how to evaluate the infinite product $$ \prod_{k = 2}^{\infty} \left( 1 - \frac{1}{k^3} \right)? $$ I know that a generalized quadratic version has a nice closed form $$ \frac{\sin(\pi z)}{\pi z}=\prod_{k=1}^\infty \left(1-\frac{z^2}{k^2}\right), $$ and wondering if something like this also exists for this product.

Unfortunately this doesn't look amenable to factorization and telescoping either, as far as I can tell, when written as $(k^3-1)/k^3$. However, I do believe it is convergent, when considered as $$ \prod_{k = 2}^{\infty} \left( 1 - a_k \right) $$ since $$ \prod_{k = 2}^{\infty} \left( 1 + a_k \right) $$ converges, though I am not an expert in this area. If there is no closed form, a numerical estimate shall suffice.

Answer 1

Let me tell you how to obtain the value from the Iosif Pinelis's answer, although I won't do the actual calculations myself and just indicate the method. The idea is as follows: $$1 - \frac{1}{k^3} = \left(1+\frac{w}{k}\right)\left(1+\frac{\bar{w}}{k}\right)\left(1+\frac{-1}{k}\right),$$

where $w = \frac{1}{2} + \frac{\sqrt{3}i}{2}$. Now, the Gamma function satisfies $$\Gamma(z) = \frac{e^{-\gamma z}}{z} \prod_{k=1}^\infty \left(1 + \frac{z}{n}\right)^{-1} e^{z/n}.$$

Thus, what we want is contained in the product $(\Gamma(w)\Gamma(\bar{w})\Gamma(-1))^{-1}$, there are just 2 caveats. First we have to observe that since $w + \bar{w} +(-1) = 0$, the exponential factors cancel in this product so we really get our desired product. The second issue is that $\Gamma(-1)$ is, umm, infinite. However, since your product starts with $k = 2$, this is not a real problem because what we are really after is $F(z)=\left(1+\frac{1}{z}\right)\Gamma(z)$, which is well-defined at $z = -1$. Specifically, since $\left(1+\frac{1}{z}\right)\Gamma(z) = \frac{1}{z^2}\Gamma(z+2)$, we get $ F(-1) = \Gamma(1) = 0! = 1$.

The last (slightly miraculous) ingredient that we need is that $w + \bar{w} = 1$ (otherwise I don't think there would've been a simpler answer than the product of Gamma factors). Therefore, the reflection formula for the Gamma function tells us that $\Gamma(w)\Gamma(\bar{w}) = \frac{\pi}{\sin \pi w}$. And now by collecting all of this, and the prefactors from the formula for the Gamma function (and knowing how to take the cosine of the complex number) we will arrive at the formula from the other answer.

Answer 2

$\newcommand\ep\varepsilon\newcommand\Ga\Gamma$Here we are going to carry out what was proposed in the previous version of the answer.

Namely, note that $$1-\frac{z^3}{k^3}=\prod_{j=0}^2\frac{k-\ep_j z}k=\frac{a_{k+1}}{a_k},$$ where $\ep_j:=e^{2\pi ij/3}$, the $j$th root $w$ of the polynomial $1-w^3$, and $$a_k=\prod_{j=0}^2\frac{\Ga(k-\ep_j z)}{\Ga(k)}.$$ So, $$\Pi(z):=\prod_{k=2}^\infty\Big(1-\frac{z^3}{k^3}\Big)= \prod_{k=2}^\infty \frac{a_{k+1}}{a_k}=\frac1{a_2}\,\lim_{n\to\infty}a_n =\frac1{a_2}=\dfrac1{\prod_{j=0}^2\Ga(2-\ep_j z)}. \tag{10}\label{10}$$ So, the product in question is $$\Pi(1)=\dfrac1{\prod_{j=0}^2\Ga(2-\ep_j)}=0.80939\ldots.$$

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More generally, this method can be used to treat any products of the form $$\prod_{k\ge k_0} P(z/k),$$ where $k_0$ is a natural number and $P$ is a polynomial.

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Details on the penultimate equality in \eqref{10}: By Striling's formula, for complex $w\ne0$ with $|\arg w|\le\pi/4$, $$\ln\Ga(w)=g(w)+O(1/|w|),$$ where $g(w):=w\ln w-w+\frac12\,\ln\frac{2\pi}w$, so that $|g''(w)|=O(\frac1{|w|}+\frac1{|w|^2})$. Using now the Taylor formula $$g(n+h)=g(n)+hg'(n)+\frac{h^2}2\int_0^1 dt\,(1-t)g''(n+th)$$ for large enough natural $n$ and $h=-\ep_j z$ with $j=0,1,2$, and noting that $\sum_{j=0}^2\ep_j=0$, we conclude that $\ln a_n=O(1/n)$ and hence $\lim_{n\to\infty}a_n=1$.

Answer 3

Thanks to Aleksei and Iosif for giving the proofs, just 1 line missing at the end using the reflection formula: result is $$\cosh(\pi\sqrt{3}/2)/(3\pi)\;.$$