# Value of $\prod_{n>1} \frac{1}{1-\frac{1}{n^s}}$ or $-\sum_{n=2}^{\infty} \log(1-\frac{1}{n^s} )$

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?

• 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 $1/2$. – Klangen Jul 16 at 9:03
• @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! – phy_math Jul 16 at 9:13
• math.stackexchange.com/q/2603561 – 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 $$k \ge 2$$ $$\prod_{n=2}^\infty (1-\frac{1}{n^k})=\prod_{m=1}^kg(e^{2i \pi m/k}), \qquad g(x) =\frac{1}{(-x)(1-x)\Gamma(-x)}= \frac{1}{\Gamma(2-x)}$$

• what is $g(x) = \frac{1}{\Gamma(-x)(-x)(1-x)}$?, you mean $\Gamma(-x)\Gamma(-x)\Gamma(1-x)$? – phy_math Jul 16 at 10:08
• @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). – Raymond Manzoni Jul 16 at 10:26
• @Raymond Manzoni Thanks! I got it – 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.