One classical Mertens' theorem tells us that $$\prod_{p \leq n} (1\frac{1}{p})^{1} = e^\gamma \log n + \mathcal{O}(1).$$ It is now very natural to ask, whether we have some good estimate to $$\prod_{p \leq n} (1\frac{1}{p^s})^{1}$$ for, let's say, $s > 1$ real. Of course the limit is $\zeta(s)$ for growing $n$, and I would like to have some portion estimate  or something similar  in the form $$\prod_{p \leq n} (1\frac{1}{p^s})^{1} = k_n \zeta(s)$$ where $k_n$ is quite well estimated with respect to $n$.

4$\begingroup$ Just take logarithms, and use partial summation (together with Mertens'estimate or the prime number theorem). $\endgroup$ – js21 May 27 '12 at 8:25

$\begingroup$ can you explain this a bit more detailed ? $\endgroup$ – tobias May 27 '12 at 9:58
I detail. It suffices to study the "tail" $P(X) = \prod_{p > X} (1 \frac{1}{p^s})^{1} $. Using $\log(1y) = y + O(y^2)$, we get for real $s>1$ $$ \log P(X) = \sum_{p > X} \frac{1}{p^s} + O \left( \frac{1}{X^{2s1}} \right) = \int_{Y>X} \frac{d\pi(Y)}{Y^s} + O \left( \frac{1}{X^{2s1}} \right) $$ $$= \int_{Y>X} \frac{dY}{Y^s \log Y} + \int_{Y>X} \frac{dR(Y)}{Y^s} + O \left( \frac{1}{X^{2s1}} \right) $$
where $R(Y) = \pi(Y)  Li(Y) $. The first integral is easy to deal with, while the second one can be treated with an integration by part (and then using known bounds of the shape $R(Y) \ll \frac{Y}{(\log Y)^A}$). For fixed $s$ this yields an asymptotic equivalent of your $k_n$.