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

The question is, how about the expression $\zeta(1) = \lim_{n \rightarrow \infty} \prod_{\textrm{primes } p_i \leq n} \frac{1}{1-p_i^{-1}}$. 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

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3 Answers 3

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Formula (8) on this page gives the result

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

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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.

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Check out

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

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