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? 

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.
