MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $n$.

share|cite|improve this question
Just take logarithms, and use partial summation (together with Mertens'estimate or the prime number theorem). – js21 May 27 '12 at 8:25
can you explain this a bit more detailed ? – tobias May 27 '12 at 9:58
up vote 8 down vote accepted

I detail. It suffices to study the "tail" $P(X) = \prod_{p > X} (1- \frac{1}{p^s})^{-1} $. Using $-\log(1-y) = y + O(y^2)$, we get for real $s>1$ $$ \log P(X) = \sum_{p > X} \frac{1}{p^s} + O \left( \frac{1}{X^{2s-1}} \right) = \int_{Y>X} \frac{d\pi(Y)}{Y^s} + O \left( \frac{1}{X^{2s-1}} \right) $$ $$= \int_{Y>X} \frac{dY}{Y^s \log Y} + \int_{Y>X} \frac{dR(Y)}{Y^s} + O \left( \frac{1}{X^{2s-1}} \right) $$

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 $k_n$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.