Infinite Product $\prod_n^\infty \frac{1}{1-\frac{1}{n^s}} \rightarrow$? Can this $$P_n(s)=\prod_{m=2}^n \frac{1}{1-\frac{1}{m^s}}$$ for s>1 and $\lim_{n\rightarrow\infty}$ be written any simpler (does it converge)? When $m$ runs here only over the primes this is the famous Euler-Product form of the Riemann zeta function. 
Is there anything known for this product when n runs over all natural numbers? For checking the convergence one could look at 
$$\log P_m(s) = -\sum_{m=2}^n \log(1-\frac{1}{m^s})$$ 
and use the integral criterion. I somehow did not succeed to integrate it.
(The idea how I came up with this was somehow to write the zeta function as quotient of this function and a similiar one with terms from the Erathostenes sieve:
$$ \zeta(s) = \frac{P(s)}{\prod_{m,n=1}^\infty \left[ 1-((m+1)(n+1))^{-s}\right]^{-\frac{1}{M(m,n)}}}.$$
Here $M(n,m)$ is obviously the number of divisors $\sigma_0$ of $nm$ minus two:
$$M(n,m)=\sigma_0(mn) -2,$$
using the prime factorisation of $nm=\prod_{i}p_i^{\alpha_i}$ it is in turn
$$\sigma_0(nm)=\prod_i (\alpha_i + 1).$$
)

Hint:
Note that $1-\frac1x = \frac{x-1}{x}$. Then you can simplify
$$\begin{align*}\log P_m(s) &= -\sum_n \log(1-\frac{1}{m^s}) \\
&= -\sum \log(\frac{m^s-1}{m^s}) \\
&= \sum\log(m^s)-\log(m^s-1) \end{align*}$$ 
