Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hi, this is my first question. It appeared while solving a research problem in cryptography. I am computer science student, so I apologize for lack of mathematical rigor in this question. Thanks for any help.

Consider the RiemannZeta function at s = +1. It diverges, but the expression for the function is RiemannZeta(1) = $\lim_{n \rightarrow \infty} \sum_{i = 1}^{n} \frac{1}{i}$ , the truncated sum of which are the $n$-th harmonic number, $\mathcal{H}(n)$.

The question is, How about the expression RiemannZeta(1) = $\lim_{n \rightarrow \infty} \prod_{\textrm{primes} p_i \leq n} \frac{1}{1-p_i^{-1}}$. is the value of the truncated product $\mathcal{H}(n)$ too?

My simulations for large values of $n$ tells me that it is some function of $\log n$ (for example comparing the ratio of the function for $n$ and $n^2$ and $n^3$ etc) How do we prove this?

In summary, What is the value of $\prod_{\textrm{primes} p_i \leq n} \frac{1}{1-p_i^{-1}}$? Thanks

share|cite|improve this question

3 Answers 3

up vote 9 down vote accepted

Formula (8) on this page gives the result

$$\prod_{p \le n} \frac1{1-p^{-1}} = e^\gamma \log n \\,(1 + o(1)).$$

share|cite|improve this answer

Note that, asymptotically, $\mathcal{H}(n)\simeq ln(n) + \gamma + \frac{1}{2n} + O(n^{-2})$. In other words, both expressions diverge like $ln(n)$ but not exactly in the same way.

share|cite|improve this answer

Check out

J. Barkley Rosser and Lowell Schoenfeld Approximate formulas for some functions of prime numbers, especially Theorem 7.

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.