I know

\begin{align} \prod_{p~is~ prime} \frac{1}{1-\frac{1}{p^2}} = \zeta(2) = \frac{\pi^2}{6} \end{align} which has a convergent number.

actually I can even generalized this to

\begin{align} \prod_{p ~is~ prime} \frac{1}{1-\frac{1}{p^s}} = \zeta(s) \end{align} For $s>1$ [consider $s\in \mathbb{R}$] we know zeta function converges, so this has a convergent number.

How about generalization to arbitrary integers? [i.e., I want to replace $p$ with arbitrary integer $n$.]

For example $s=2$, we have

\begin{align} \prod_{n>1} \frac{1}{1-\frac{1}{n^2}} \end{align}

taking log we need to show \begin{align} - \sum_{n=2} \log\left(1-\frac{1}{n^2} \right) \end{align} is convergent or not.

simply by telescope method I can see this value converges to $\log(2)$, that means $ \prod_{n>1} \frac{1}{1-\frac{1}{n^2}} = 2$.

Now consider $s>1$.

\begin{align} -\sum_{n>1} \log\left(1-\frac{1}{n^s}\right) \end{align} This is convergent from comparison test.

Simply take $a_n = -\log(1-\frac{1}{n^s})$ and $b_n = \frac{1}{n^s}$, then \begin{align} \lim_{n\rightarrow \infty} \frac{a_n}{b_n} = \lim_{x\rightarrow 0} \frac{-\log(1-x)}{x} = 1 >0 \end{align} and since $\sum_{n=1}^{\infty} b_n = \zeta(s)$ is convergent for $s>1$, $\sum_{n=2}^{\infty} a_n$ also converges.

What I want to obtain is the value of such convergent series, first i tried telescope method, but it seems difficult even for $s=3$.

Is there a way to compute exact value of those products?

How and what is the values of those products?

  • 1
    $\begingroup$ As far as I'm concerned no closed form formula is known for $s>2$. By the way, when $s=2$ the product equals $2$, not $1/2$. $\endgroup$ – Klangen Jul 16 at 9:03
  • 2
    $\begingroup$ @Klangen, thanks, I found typos and corrected it! $ \sum_{n=2}^{\infty} \log(1-\frac{1}{n^2} ) = - \log(2)$. I am quite surprised that for $s>2$, no closed form formula is known! $\endgroup$ – phy_math Jul 16 at 9:13
  • 1
    $\begingroup$ math.stackexchange.com/q/2603561 $\endgroup$ – user514787 Jul 16 at 9:53

For $2k\ge 2$ $$\prod_{n=2}^\infty (1-\frac{1}{n^{2k}})=\prod_{m=1}^{2k} \prod_{n=2}^\infty (1-\frac{e^{2i \pi m/(2k)}}{n}) = \prod_{m=1}^{k}\prod_{n =-\infty, |n|\ge 2}^\infty (1-\frac{e^{2i \pi m/(2k)}}{n}) = \prod_{m=1}^{k} f(e^{2i \pi m/(2k)})$$ where $f(x) =\frac{\sin(\pi x)}{\pi x(1-x^2)} $ and in those products the order of summation is meant to be $\lim_{N \to \infty} \prod_{|n| \le N} $

For $k \ge 2$ $$\prod_{n=2}^\infty (1-\frac{1}{n^k})=\prod_{m=1}^kg(e^{2i \pi m/k}), \qquad g(x) =\frac{1}{(-x)(1-x)\Gamma(-x)}= \frac{1}{\Gamma(2-x)}$$

  • $\begingroup$ what is $g(x) = \frac{1}{\Gamma(-x)(-x)(1-x)}$?, you mean $\Gamma(-x)\Gamma(-x)\Gamma(1-x)$? $\endgroup$ – phy_math Jul 16 at 10:08
  • $\begingroup$ @phy_math: he means $\;g(x)=\dfrac{1}{x(x-1)\Gamma(-x)}\;$ which returns the correct result (notice that he is computing the multiplicative inverse of your formula). $\endgroup$ – Raymond Manzoni Jul 16 at 10:26
  • $\begingroup$ @Raymond Manzoni Thanks! I got it $\endgroup$ – phy_math Jul 16 at 10:38

Let $$P_s=\prod_{n=2}^\infty \frac{1}{1-n^{-s}}$$

Using a CAS, there are some nice expressions such as $$\left( \begin{array}{cc} s & P_s \\ 2 & 2 \\ 3 & 3 \pi \text{sech}\left(\frac{\sqrt{3} \pi }{2}\right) \\ 4 & 4 \pi \text{csch}(\pi ) \\ 6 & 6 \pi ^2 \text{sech}^2\left(\frac{\sqrt{3} \pi }{2}\right) \end{array} \right)$$ $P_5$ and the other ones are quite ugly.

Have a look here.


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