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
$$\sigma_0(nm)=\prod_i (\alpha_i + 1).$$
)
 A: As noted in the comments since $\sum \frac{1}{n^k}$ converges for $k>1$ so does the product. I will here derive a formula for $P(k)$ that is valid for integers $k>1$. We start by factoring 
$$\frac{1}{1-\frac{1}{n^k}} = \frac{1}{\left(1-\frac{1}{n}\right)\left(1-\frac{e^{\frac{2\pi i}{k}}}{n}\right)\cdots \left(1-\frac{e^{\frac{2\pi i(k-1)}{k}}}{n}\right)} =  \prod_{i=0}^{k-1}\left(1+\frac{z_k^i}{n}\right)^{-1}$$ where $z_k = -e^{\frac{2\pi i}{k}}$. Changing the order of the products gives us
$$P_N(k) = (-1)^{k+1}\prod_{i=0}^{k-1}\frac{e^{-\gamma z_i}}{z_i}\prod_{n=2}^N\left(1+\frac{z_k^i}{n}\right)^{-1}e^{\frac{z_i}{n}}$$
where we have used $z_0z_1\cdots z_{k-1}=(-1)^{k+1}$ and $z_0+z_1+\ldots+z_{k-1} = 0$ which implies $\prod_{i=0}^{k-1}e^{-\gamma z_i} = \prod_{i=0}^{k-1}e^{\frac{z_i}{n}} = 1$ and $\prod_{i=0}^{k-1}z_j = (-1)^{(k+1)}$. 
These terms have been inserted for the next step which is to use the Weierstrass product formula for $\Gamma(z)$ which allows us to write
$$P_N(k) = (-1)^{k+1}\prod_{i=0}^{k-1}\Gamma_N(z_i)(1+z_i)$$
where $\lim_{N\to\infty}\Gamma_N(z) = \Gamma(z)$. Since $\lim_{z\to -1}\Gamma(z)(1+z) = 1$ the $i=0$ term does not contibute to the product (the treatement of the $i=0$ term should be done more carefully above; but the naive treatement here works) and by taking $N\to\infty$ and using $\Gamma(2+z_i)=(1+z_i)z_i\Gamma(z_i)$ we obtain
$$P(k) = \prod_{i=1}^{k-1}\Gamma(2+z_i) = \prod_{j=1}^{k-1}\Gamma\left(2 - e^{\frac{2\pi i j}{k}}\right)$$
In particular this gives the values
$$P(2) = 2,~~~P(3) = \frac{3\pi}{\cosh\left(\frac{\sqrt{3}\pi}{2}\right)},~~~P(4) = \frac{4\pi}{\sinh(\pi)}$$
and as $k\to\infty$ we have the approximation $P(k) \simeq 1 + \frac{1}{2^k}$.
A: Hint:
Note that $1-\frac1x = \frac{x-1}{x}$. Then you can simplify
$$\begin{align*}\log P_m(s) &= -\sum_n \log(1-\frac{1}{m^s}) \\
&= -\sum \log(\frac{m^s-1}{m^s}) \\
&= \sum\log(m^s)-\log(m^s-1) \end{align*}$$ 
